How can space be euclidean when light bends? I have read people arguing that tridimensional space sections of space time continuum (whatever its number of dimensions) appears to be euclidean from empirical evidence. I cannot reconcile it with my understanding that


*

*Mass does exist 

*Mass does curve the spacetime continuum and 3D space sections 

*light bending has been explained by spacetime curvature. 

 A: It takes time for light to bend, i.e. the bending of light is a consequence of the curvature of geodesics in four dimensional space-time. On the other hand, a three dimensional snapshot of space appears to have geodesics that are straight lines (some specifics shown below). This is exactly why we thought space was Euclidean for so long and why all Earth-bound experiments on a local scale mesh well with the hypothesis of Euclidean geometry.
Take, for example, the Schwarzschild metric of space-time around a spherical mass (taken from Wikipedia):
$$
ds^2 = \left( 1 - \frac{r_s}{r} \right) dt^2 - \left( 1 - \frac{r_s}{r} \right)^{-1} dr^2 - r^2 \left( d \theta^2 + \sin^2 \theta d \phi^2 \right)
$$
For $r \gg r_s$, the spatial component of this metric is simply
$$
dl^2 = -ds^2 \mid_{dt = 0} = dr^2 + r^2 \left( d \theta^2 + \sin^2 \theta d \phi^2 \right)
$$
which is just the 3-dimensional Euclidean metric in spherical coordinates. Of course, this is only approximately Euclidean (and the approximation breaks down for $r \sim r_s$), but it was enough to fool us all until Einstein.
So, this should be pretty easy to reconcile with empirical evidence. Just go look at things, and ask yourself "High school geometry works pretty accurately, doesn't it?".
A: The statement that space is Euclidean is a broad statement, not meant to hold near very massive bodies or in arbitary volumes. One sense in which it can be meant is to hold "on average" for the whole spacetime - the universe - as such.
The best candidate for the overall metric of spacetime is the FRLW metric, which is the exact solution for a universe that is homogenous and isotropic (looks the same at all points in all directions). On large scales, the homogeneity of the Cosmic Microwave Background suggests this is a good approximation to our universe at large. It is written as
$$ \mathrm{d}s^2 =  - \mathrm{d}t^2 + a(t)^2\mathrm{d}\Sigma^2$$
modulo an overall sign and a factor of $c$ set to $1$, where $\mathrm{d}\Sigma^2$ is the metric of a three-dimensional space of uniform curvature. Now, there are three different kinds of space that could be - a hyperbolic (negative curvature), an elliptic (positive curvature) or an Euclidean (flat) space. 
These are, effectively, distinguished by the way the angles in a triangle add up. Now, look at the space around you, on astronomical scales (so take a triangle out of stars). Observations by astronomers in this vein have found that there's no major indication that the space part of spacetime is curved on big scales (though locally, it may well be), though, due to experimental errors, we can of course only say that the curvature is very close to zero.
Another way to get to Euclidean space is to make a weak field approximation, i.e. being far from the Schwarzschild radius of a supermassive body.
A: You are absolutely correct that three-dimensional sections of space-time do not satisfy Euclidean geometry -- they are not flat. However, they are almost flat. On room-sized scales the curvature is very small indeed.
I don't know the exact context where you read that "tridimensional space sections of space time continuum (whatever its number of dimensions) appear to be Euclidean from empirical evidence," but I assume the writers were not claiming that spatial sections truly are Euclidean (in the sense of being flat), to the best of our understanding and to infinite precision. They only meant that such sections appear to be Euclidean when observed using simple methods.
