Removed by the author
It is certainly true for non-singular matrices $\displaystyle A$ (as the corresponding linear map has a linear (hence continuous) inverse).
I'm not sure about the singular matrices. It seems very likely that, in general, singular matrices are closed maps in the sense that they map closed sets to closed sets.
We do have the isomorphism
$\displaystyle \mathbb{R}/\text{Ker}(A)\simeq \text{Im}(A),$
which gives us a factorization of $\displaystyle A$ (as a linear operator) as the composition of a projection $\displaystyle P\colon \mathbb{R}^n\rightarrow\mathbb{R}^n/\text{Ker}(A)$ composed with the inclusion of $\displaystyle \mathbb{R}^n/\text{Ker}(A)\simeq\text{Im}(A)$ into the codomain $\displaystyle \mathbb{R}^m$.
The inclusion is certainly a closed map, and so it suffices to check that projections are also closed maps.
Ugh, it is clearly wrong that all singular matrices are closed maps. I must have been too focused on it being true that I didn't care to look for an easy counterexample. You can take the projection $\displaystyle (x,y)\mapsto x$, for which the closed set given by the graph of $\displaystyle f(x) = 1/x$ maps to an open set of $\displaystyle \mathbb{R}$.
For this particular problem, I asked around, and someone came up with this idea:
The set $\displaystyle X$ is the cone of the unit cube $\displaystyle [0,1]^n$, i.e. $\displaystyle X=C([0,1]^n)$, where $\displaystyle C$ is the cone operator, that is, $\displaystyle C(S) = \{\lambda a | \lambda\geq 0, a\in S\}$. For any linear operator $\displaystyle A$, and any set $\displaystyle S$, it is true that
$\displaystyle A(C(S)) = C(A(S)).$
Hence $\displaystyle A(X) = C(A([0,1]^n))$, which is the cone of a compact set (since all linear maps are continuous, even the singular ones), but the cone of a compact set is closed.
I've also recalled that if $\displaystyle Y$ is a compact topological space, then the projection $\displaystyle \pi_X\colon X\times Y\rightarrow X$ is a closed map - just for bonus info.