In a topological vector space, why does $M$ closed and $Y$ finite dimensional imply that $M$ is topologically complemented in $M+Y$ Let $X$ be a topological vector space. If $Y$ and $M$ are subspaces such that $Y$ is finite dimensional and $M$ is closed why is $M$ topologically complemented in $M+Y$ (i.e. why is $A:M\times Y\rightarrow M+Y$ defined by $A(m,y) = m+y$ a homeomorphism.)
I have shown that $Y$ being finite dimensional implies that $Y$ is closed in $X$ and also that $Y+M$ is closed in $X$. It is clear that $A$ is a continuous bijection so it only remains to show that its inverse is continuous. For this reason I want to argue by using nets. If $\{y_i\}$ and $\{m_i\}$ are nets in $Y$ and $M$ respectively such that $y_i+m_i\rightarrow z$ then since $Y+M$ is closed $y_i+m_i\rightarrow y+m$. If I could show that $y_i\rightarrow y$ and $m_i\rightarrow m$ I would be done, but I don't see how to prove this and would be thankful for any tip.

Edit: Maybe one could argue as follows the natural map $Q:Y+M\rightarrow (Y+M)/M$ is continuous and the dimension of $(Y+M)/M$ is the same as the dimension of $Y$. Therefore we can find a continuous map from $Y+M$ to $(Y+M)/M$ and then to $Y$. Denoting the map with $B$ we find by continuity that $y = B(y+m) = \lim B(y_i+m_i) = \lim B(y_i) = y_i$
 A: Your idea is correct.
Let $Q : M + Y \to Y' = (M+Y)/M$ be the quotient map and endow $Y'$ with the quotient topology. Then $Y'$ is a topological vector space (TVS). Moreover, since $M$ is closed in $M + Y$, the TVS $Y'$ is Hausdorff. That it is the essential ingredient. 
It is well-known that on a finite-dimensional vector space  $V$ there exists a unique Hausdorff topology making $V$ a TVS (it is induced by any norm on $V$). Thus all linear maps between finite-dimensional Hausdorff TVS's are continuous.
The linear map $\phi = Q \mid_Y : Y \to Y'$ is clearly a linear isomorphism. Hence it is a homeomorphism. Now consider $\psi = \phi^{-1} \circ Q : M + Y \to Y$. This is a continuous linear map. For $m + y \in M  + Y$ we have $\psi(m + y) = \phi^{-1}(Q(m+y)) = \phi^{-1}([y]) = y$. Also the map $\theta : M + Y \to M + Y, \theta(x) = x - \psi(x)$, is continuous. We have $\theta(m+y) = m+y - y = m$, thus we may regard $\theta$ as a map into $M$ which is clearly continuous. This shows that
$$f : M + Y \to M \times Y, f(m+y) = (m,y),$$
is continuous.
