Whether an in-falling observer can see (or more generally be able to receive sensory input from) their feet as they pass through the event horizon is an interesting question. The answer is yes - you will be able to see your feet the entire time, putting aside the whole spaghettification issue.
This question provides an excellent opportunity to examine what coordinates actually mean in GR, and how we define and interpret them. If you want the punchline, skip to the end of the answer.
The reason can best be understood by adopting the Gullstrand-Painlevé coordinates, also called raindrop coordinates. In this coordinate system, we utilize the same $r,\theta,\phi$ coordinates as the standard Schwarzschild metric along with a new time coordinate
$$T \equiv t - 2M\left(-2y + \log\left[\frac{y+1}{y-1}\right]\right), \qquad y\equiv \sqrt{\frac{2M}{r}}$$
in units such that $c=G=1$.
Before proceeding, it is worth taking a moment to review the precise physical interpretation of the Schwarzschild coordinates.
Because the metric is static, we may slice the spacetime into space-like surfaces $\Sigma_t$ labeled by a timelike coordinate $t$ in such a way that
- the components of the spatial metric $g_\Sigma$ on $\Sigma_t$ are independent of $t$, and
- the vector field $\partial_t$ is orthogonal to the surfaces $\Sigma_t$.
With this choice of time coordinate, we manifest the idea that the spacetime is not changing with time; any two slices $\Sigma_t$ and $\Sigma_{t'}$ are identical (more precisely, isometric). Furthermore, if an object is at rest on the slice $\Sigma_t$ (by which I mean that $\mathrm dx^i/dt = 0$ for each of the spatial coordinates $x^i$), then it will have the same spatial coordinates on the slice $\Sigma_{t+\delta t}$ (to first order in $\delta t$).
Because the metric is spherically symmetric, each $\Sigma_t$ can be decomposed into a family of concentric $2$-spheres. More precisely, given a point $p\in \Sigma_t$, the set of points which can be reached from $p$ by performing a rotation about the center of the black hole constitutes a 2-sphere, which can be coordinatized with the standard angles $(\theta,\phi)$. Spherical symmetry implies that the induced metric on each sphere is given by $g_{\mathrm S^2} = K\big(\mathrm d\theta^2 + \sin^2(\theta) \mathrm d\phi^2\big)$, where $K\in(0,\infty)$ does not depend on $\theta$ or $\phi$.
The surface area of each of the aforementioned 2-spheres is given by
$$A = \int_0^{2\pi} \mathrm d\phi \int_0^\pi \mathrm d\theta \sqrt{\mathrm{det}(g_{\mathrm S^2})}=4\pi K$$
If we define the coordinate $r \equiv \sqrt{K} = \sqrt{A/4\pi}$, then the metric on $\Sigma_t$ takes the form
$$g_{\Sigma} = a(r)\mathrm dr^2 + r^2\big(\mathrm d\theta^2 + \sin^2(\theta)\mathrm d\phi^2\big)$$
where $a$ is an unknown function.
Under these assumptions, the full spacetime metric $g$ is given by
$$g = -b(r) \mathrm dt^2 + a(r) \mathrm dr^2 + r^2 \big(\mathrm d\theta^2 + \sin^2(\theta) \mathrm d\phi^2\big)$$
where $a$ and $b$ are unknown functions. Plugging them into the Einstein equations and applying the appropriate boundary conditions (specifically, that the spacetime approaches a Newtonian limit at $r\rightarrow \infty$), we find that
$$a(r) = \frac{1}{b(r)} = 1- \frac{2M}{r}$$
where $M$ is the mass of the black hole. The full metric in Schwarzschild coordinates is then
$$g = -\left(1-\frac{2M}{r}\right)\mathrm dt^2 + \left(1-\frac{2M}{r}\right)^{-1}\mathrm dr^2 + r^2\big(\mathrm d\theta^2 + \sin^2(\theta) \mathrm d\phi^2\big)\tag{$\star$}$$
So to conclude this review:
- The temporal component $t$ is chosen to make the metric components independent of $t$, and the spacelike surfaces of simultaneity $\Sigma_t$ orthogonal to the "time direction" $\partial_t$. If we would like to assign a physical meaning to the coordinate $t$, we might imagine a family of observers each of whom sits at fixed spatial coordinates $(r,\theta,\phi)$, and who all start their wristwatches when their worldlines intersect the surface $\Sigma_0$. When we say that an event has coordinates $(t,r,\theta,\phi)$, we mean that it occurs at spatial position $(r,\theta,\phi)\in \Sigma_t$ and that as we walk away from the black hole along the slice $\Sigma_t$, the time on the observers' wristwatches will approach $t$ as $r\rightarrow \infty$. As a crucial note, $t$ is not the time on the wristwatch of the observer fixed to the point $(r,\theta,\phi)$; their watch reads $t\sqrt{1-\frac{2M}{r}} < t$.
- The angular coordinates $(\theta,\phi)$ are chosen because we decompose each slice $\Sigma_t$ into a family of nested 2-spheres.
- We label each such 2-sphere by the areal radius $r\equiv \sqrt{A/4\pi}$. Note that this is not this distance from a point on the 2-sphere to the center of the black hole! Rather, it is the square root of the area of the 2-sphere divided by $4\pi$. This coincides with the radial distance if $\Sigma_t$ is flat - but that is not the case here.
With that review under our belts, we may now provide an interpretation to the raindrop time $T$. Rather than choosing our time coordinate to reflect the static nature of the metric as we did with $t$, we should like to choose a time coordinate which reflects the experience of a radially in-falling observer. To do this, we first draw the trajectories of in-falling observers in Schwarzschild coordinates, where each observer starts from some initial radius $r_0$. We will take $r_0$ sufficiently large that the gravitational time dilation is negligible, and so the time coordinate $t$ corresponds to the wristwatch time of a stationary observer fixed at $r_0$.
A generic event $E$ lies on exactly one of these trajectories. When the in-falling observer departs from $r_0$ at Schwarzschild time $t=t_0$, their wristwatch time matches the Schwarzschild time. When they arrive at $r_E$, however, their watch is no longer synchronized with the observer at $r_E$; this is due to a combination of gravitational and kinematic time dilation.
One can compute the discrepancy between the wristwatch time of our in-falling observer and the local Schwarzschild time at $t_E$. Between the time the observer departs from $r_0$ and the time they arrive at $r_E$, the time elapsed on their wristwatch $\Delta \tau$ differs from the elapsed Schwarzschild time $\delta t \equiv t_E-t_0$ by an amount
$$\Delta \tau - \Delta t = 2M \Delta\left(2y + \log\left[\frac{y+1}{y-1}\right]\right), \qquad y \equiv \sqrt{\frac{2M}{r}}$$
which is independent of $r_0$ except for the fact that we have sneakily used the assumption that $r_0 \gg 2M$ during the derivation (figuring out where is a good exercise).
This provides us with the desired interpretation - the coordinate $T$ of an event is the wristwatch time of an in-falling observer, whose clock was initially synchronized to the Schwarzschild time but has lost synchronicity due to kinematic and gravitational time dilation.
In these new raindrop coordinates, the spacetime metric takes the form
$$g = -\left(1-\frac{2M}{r}\right) \mathrm dT^2 - 2\sqrt{\frac{2M}{r}} \mathrm dT \mathrm dr + \mathrm dr^2 + r^2 \big(\mathrm d\theta^2 + \sin^2(\theta) \mathrm d\phi^2\big)$$
This differs from $(\star)$ in two major ways. First, this expression is not singular at $r=2M$, and so it covers the entire Schwarzschild spacetime (not just the region outside of the event horizon). Second, the spacelike hypersurfaces of simultaneity $\Sigma_T$ are now spatially flat (the spatial part of the metric looks like ordinary spherical coordinates); the price we pay for this is the cross term proportional to $\mathrm dT\mathrm dr$, which we can loosely think of as as a kind of drag - a natural "flow" through space towards $r=0$. See e.g. the river model of black holes for a more thorough description of this picture.
In these coordinates, the velocity of an in-falling observer (our namesake raindrop) who was initially at rest at $r_0\rightarrow \infty$ is found to be
$$\frac{dr_{raindrop}}{dT} = - \sqrt{\frac{2M}{r}}$$
On the other hand, the trajectories of light beams are given by
$$\frac{dr_{light}}{dT} = - \sqrt{\frac{2M}{r}} \pm 1$$
We see that inside the event horizon where $r<2M$, light beams must travel toward the center - $dr/dT <0$ - but can do so at two different rates, corresponding to us pointing our flashlights toward or away from $r=0$. Even in the latter case, the light travels only radially inward, but it does so more slowly than our freely-falling observer.
The Punchline
In the context of this question, once you get inside the event horizon the light emanating from the tops of your shoes travels inward toward $r=0$ at a rate of $dr/dT = -\sqrt{2M/r} + 1$, which is smaller (i.e. less negative) than the rate at which your head falls toward $r=0$, namely $dr/dT = -\sqrt{2M/r}$. The relative speed between the light and your eyes is then simply $1$ (or $c$, in SI units, as we might expect from the equivalence principle), and so you will indeed be able to see whether your shoes remain tied as you are swallowed by the gaping maw of infinity.
