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Schwarzchild metric is static, none of its components change over time. Curiously, basis vector $e_r$ does change over time. If we differentiate $e_r$ with respect to time and take the time component of the derivative we get a non zero value, i.e, $\Gamma_{10}^0$ is not $0$.
So, if $e_r$ changes over time, how come $g_{rr}$ remain static, given that $g_{rr} = e_r\cdot e_r$?
We need to quantify the meaning of "changes with time". There's two ways to define changes of a vector field with respect to another one. One of them is given by the covariant derivative, which you have calculated:
$$ \nabla_{\partial_{t}}\partial_{r} = \Gamma^{\lambda}_{tr}\partial_{\lambda}$$
and have correctly noticed that only $\lambda=t$ contributes to the above, yielding:
$$ \nabla_{\partial_{t}}\partial_{r} = \frac{1}{2}g^{tt}g_{tt,r}\partial_{t}$$
However, this does not translate in any way to the claim that $g_{rr}$ changes with time. The only thing it says is that the coordinate vector field $\partial_{r}$ is not geodesic (c.f. the meaning of the parallel transport and geodesics).
The other way of quantifying the change of a vector field with respect to another one is through Lie derivative:
$$ \mathcal{L}_{\partial_{t}}\partial_{r} = [\partial_{t},\partial_{r}] = 0$$
Which means that the coordinate vector field is Lie-transported along $\partial_{t}$ (transported along the integral curves of one-parameter diffeomorphism group of the vector field $\partial_{t}$, $\mathcal{G}^{\partial_{t}}$).
Note that this still doesn't necessarily mean that the metric component changes with time. The only way to meaningfully define the metric's rate of change with time is to calculate the Lie derivative:
$$ \mathcal{L}_{\partial_{t}}g = \mathcal{L}_{\partial_{t}}(g_{\mu\nu}dx^{\mu}\otimes dx^{\nu})$$
which you can find to vanish.
A better answer
We're posing the question: how does the four length of the $\partial_{r}$ coordinate vector field change in time?
$$f := g_{rr}=g(\partial_{r},\partial_{r}) = (1 - \frac{2M}{r})^{-1}$$
Since the above is a scalar field, to calculate its derivative along a vector field, we can equivalently use:
$$ \partial_{t}f = \mathcal{L}_{\partial_{t}}f = \nabla_{\partial_{t}}f$$
The first equivalent formula
In the first case one we calculate trivially
$$ \partial_{t}f = \partial_{t}\big{(}(1 - \frac{2M}{r})^{-1}\big{)} = 0 $$
This should already settle the matter, but we can calculate the other two as an exercise. In the second and the third one, we use the commutativity of the covariant/Lie derivative with contractions.
The second equivalent formula
$$ \mathcal{L}_{\partial_{t}}f = \mathcal{L}_{\partial_{t}}(g(\partial_{r},\partial_{r})) = (\mathcal{L}_{\partial_{t}}g)(\partial_{r},\partial_{r}) + g(\mathcal{L}_{\partial_{t}}\partial_{r}, \partial_{r}) + g(\partial_{r},\mathcal{L}_{\partial_{t}}\partial_{r})$$
The first term is what I've mentioned before - the metric tensor is Lie-transported along the flow of $\partial_{t}$. If you need help proving that, here's a quick calculation, using the fact that the Lie derivative (also the covariant derivative) obeys the Leibniz rule:
$$\mathcal{L}_{\partial_{t}}g= \mathcal{L}_{\partial_{t}}(g_{\mu\nu}dx^{\mu}\otimes dx^{\nu}) = \mathcal{L}_{\partial_{t}}(g_{\mu\nu})dx^{\mu}\otimes dx^{\nu} + g_{\mu\nu}(\mathcal{L}_{\partial_{t}}dx^{\mu})\otimes dx^{\nu} + g_{\mu\nu}dx^{\mu}\otimes (\mathcal{L}_{\partial_{t}}dx^{\nu}) = 0 $$
It vanishes because: $\mathcal{L}_{\partial_{t}}(g_{\mu\nu}) = \partial_{t}g_{\mu\nu}=0$
where the Lie derivative reduces to a partial derivative because it acts on a scalar field.
Meanwhile the other Lie derivatives can be evaluated by virtue of Cartan's magic formula:
$$ \mathcal{L}_{X}\omega = d(\iota_{X}\omega) + \iota_{X}d\omega $$
$$ \mathcal{L}_{\partial_{t}}dx^{\mu} = d(dx^{\mu}(\partial_{t})) + \iota_{\partial_{t}}d^{2}x^{\mu} = d(\delta^{\mu}_{t}) = 0 $$
The third equivalent formula
$$\nabla_{\partial_{t}}f = \nabla_{\partial_{t}}(g(\partial_{r},\partial_{r})) = (\nabla_{\partial_{t}}g)(\partial_{r},\partial_{r}) + g(\nabla_{\partial_{t}}\partial_{r}, \partial_{r}) + g(\partial_{r},\nabla_{\partial_{t}}\partial_{r}) = 0 $$
where we use the metric compatibility of the covariant derivative and the fact that $\nabla_{\partial_{t}}\partial_{r} = \frac{1}{2}g^{tt}g_{tt,r}\partial_{t} $ and $\partial_{t}$ is orthogonal to $\partial_{r}$.
I think I got this. $$g_{rr} = e_r\cdot e_r$$ $$\Rightarrow \partial_tg_{rr} =2e_r\cdot\partial_te_r$$ Now, since the time derivative of $e_r$ is solely in the direction of $t$, it is orthogonal to $e_r$, therefore the RHS is $0$, equating the LHS.
