Closed plus finite dimensional in a TVS If $E$ is a topological vector space (TVS), $F_1$ a closed subspace of $E$, and $F_2$ a finite dimensional subspace of $E$, such that $F_1 \cap F_2=\{0\}$, is $F_1+F_2$ necessarily closed? If yes, are the projection from $F_1+F_2$ onto $F_1$ and $F_2$ respectively, continuous? 
 A: Yes If we suppose that $E$ is a normed space , the sum of a closed subspace $Y$ and a finite dimensional space $F$ is closed, in fact :
suppose that a sequence $(y_n+f_n) \subset Y+F$ converge vers $x\in X$, Let $P$ the (continuous Why? ) projection from $Y+F$ to $F$, the sequence $(y_n+f_n)$ is bounded (because it converge to $x$ so it exist a subsequence $(f_{n_k})$ that converge to $f\in F$ (because F is finite dimensional vector space so every bounded closed set is compact) and so $y_{n_k}$ converge to $y\in Y$ as difference of two convergent sequences, and so $x=y+f\in Y+F$
To complete the proof we need to proof that the  projection from $Y+F$ to $F$ is continuous, so for that we put 
$$
\delta=\min\{d(f,Y) : \|f\|=1 \}
$$
$\delta>0$ because the set $S_F=\{ f\in F ; \|f\|=1\}$ is compact, and the function $d(.,Y)$ is a continuous function so it exist $f'\in S_F$ such that $\delta=d(f',Y)$, then if $\delta=0$ this implies that $f'\in Y\cap F=\{0\}$ but $\|f'\|=1\neq 0$, absurde. 
so 
$$
\|y+f\|\geq \delta\|f\|
$$
so the projection from $Y+F$ to $F$ is of norme $\leq \delta^{-1}$
and the projection from $Y+F$ to $Y$ is of norme $\leq 1+\delta^{-1}$
A: Whenever N,F are subspaces of a TVS, N closed and F finite-dimensional, N+F is closed.
This is Theorem 1.42 (p. 30) in Rudin: Functional Analysis, 2nd edition, 1991.
https://archive.org/details/RudinW.FunctionalAnalysis2e1991
But even if the TVS is Hilbert, F being closed is not enough (By Exercise 1.20, p. 40), it must be finite-dimensional.
Theorem 5.16b (p. 126):
If N and F are closed subspaces of an F-space N+F, and $N\cap F=\{0\}$, then the projection $P:A+B\to A$ with null space $B$ is continuous.
So to your first question the answer is "yes", to the second "yes if N+F is an F-space" but not in general, by Exercise 5.9.
Exercise 5.9 (p. 145): Here $N,F$ are closed subspaces over $L^2(0,1)$, $N\cap F=\{0\}$ but yet $P$ is not continuous.
Note: it is not enough that N,F are closed subspaces of the F-space $L^2$; the space $N+F$ should be complete (F-space).
Def. 1.8e: F-space = TVS whose topology is induced by a complete invariant metric (equivalently, TVS having a complete N1 topology, by Theorem 1.24)
BTW, only in Theorem 1.42 finite-dimensionality is assumed, not in the others. 
In Rudin, TVSs are Hausdorff.
