I would post this as a comment, since I need to check it a little more before I am totally ok with it--alas, it is too long. So, an answer it shall stay.
EDIT: I should emphasize that where I am using finite dimesionality is that every linear map in sight is continuous. In particular, a linear isomorphism is a homeomorphism.
I think this should suffice. Let $V$ and $W$ be finite dimensional real vector spaces and $T:V\to W$ a surjective linear map.
EDIT EDIT: As Matt E points out below, it is prudent to mention the following. Below I prove that the map $\pi$ is open when we give $V/Y$ the quotient topology, but not that it is open when we give it the standard metric topology which makes $\widetilde{T}$ a homeomorphism. But, the fact that the quotient topology on $V/Y$ coincides with the linear topology is deducible because of the following: a) we're in finite dimensions so $Y$ is closed, and b) give $V$ you're favorite norm, since $Y$ is closed it's ommonly known then that the quotient topology on $V/Y$ is the topology induced by the quotient norm and so the quotient topology is the linear topology (all normed topologies are the same in finite dimensions). There are other ways to see that the two topologies coincide if you're interested. :)
We first claim that the projection map $V\to V/Y$ is open for any $Y$ a subspace of $V$. To see this, we note that it suffices to prove that $\pi^{-1}(\pi(U))$ (where $\pi$ is the projection map) is open for each open set $U$ in $V$. But, note that $\pi^{-1}(\pi(U))=\bigcup_{v\in Y}(v+U)$
which, being the union of open sets, is open.
Now that we have the projection map is open this should be easy. Namely, let $Y=\ker T$. We know then that $T$ induces a linear isomorphism $\widetilde{T}:V/Y\to W$. Since this is a linear isomorphism, it is, in particular, a homeomorphism and so an open map. So, if we take $U\subseteq V$ open we have that $T(U)$ is just $\widetilde{T}(\pi(U))$ where $\pi$ is the projection $V\to V/Y$. So, we see that $\pi(U)$ is open in $V/Y$ since $\pi$ is an open map, and $\widetilde{T}(\pi(U))$ is open because $\widetilde{T}$ is a homeomorphism. So, $T(U)=\widetilde{T}(\pi(U))$ is open as desired.