A surjective linear map into a finite dimensional space is open I'm in search of different proofs of the following proposition:

$\bf{Proposition}$: Suppose $X$ and $Y$ be topological vector spaces, $\text{dim }Y<\infty$, and $\Lambda:X\to Y$ is a surjective linear map. Then $\Lambda$ is open.

Any and all proofs are welcomed.
 A: Thanks to Danielsen for catching an error in my previous proof.
How about this approach: Let $e_i$ be basis vectors of $\mathbb{K}^n$ (ie, $Y$), and choose $x_i \in X$ such that $\Lambda x_i = e_i$. Now consider the map $\phi: \mathbb{K}^n \to X$ given by $\phi(\alpha) = \sum \alpha_i x_i$. $\phi$ is continuous since $X$ is a tvs. Also, we note that $\Lambda \circ \phi$ is the identity mapping.
Suppose $U \subset X$, then since $\phi (\phi^{-1} U) \subset U$, we see that $\phi^{-1} U = \Lambda \circ \phi (\phi^{-1} U) \subset \Lambda U$.
Let $U \subset X$ be an open neighbourhood of $0 \in X$. Then, by continuity, $\phi^{-1} U$ is open, $0 \in \phi^{-1} U$, and hence $\Lambda U$ contains an open neighbourhood of $0 \in Y$.
Now suppose $U \subset X$ is open, and $y_0 \in \Lambda U$. Then $y_0 = \Lambda x_0$ for some $x_0 \in U$. Let $U' = U -\{x_0\}$, which is an open neighbourhood of $0$. Then $\Lambda U'$ contains an open neighbourhood $V' \subset Y$ of $0$. Then $V = V'+\{\Lambda x_0\} = V' + \{y_0\} \subset \Lambda U$ is an open neighbourhood of $y_0$. Hence $\Lambda$ is an open map.
A: I don't know what topological vector spaces are. But what I know is that any finite dimensional real vector space can be made into a topological space by choosing any norm on it which does not depend on the choice of the norm.
Now if $V$ is any finite dimensional real vector space and $W$ is a linear subspace of $V$, the projection map $\pi:V\to V/W$ can be easily verified to be an open map.
Any surjective linear map $T:V\to Y$ factors uniquely through a linear isomorphism $\bar T:V/\ker T\to Y$.
Thus $T$ is a composition of two open maps and thus itself is open.
Hope this at least throws some light on the problem.
