Let $E$ be a Banach space and let $S$ and $T$ be closed subspaces, with dim$\space T<\infty$. Prove that $S+T$ is closed.
To prove that $S+T$ is closed I have to show that for any limit point $x$ of $S+T$, $x \in S+T$. So let $x$ be a limit point that subspace. Then, there is a sequence $\{x_n\}_{n \in \mathbb N} \subset S+T$ : $x_n \rightarrow x$. $x_n \in S+T \space \forall \space n \in \mathbb N$. This means that $x_n=s_n+t_n$ with $s_n \in S$ and $t_n \in T$ $\forall \space n \in \mathbb N$. Well, I couldn't go further than this. $\{s_n\}_{n \in \mathbb N}$ and $\{t_n\}_{n \in \mathbb N}$ are sequences in $S$ and $T$ respectively. I would like to prove that both of them converge to some points $s$ and $t$, for if this is the case, the hypothesis would assure $s \in S$ and $t \in T$. Then, I would have to prove that $x=lim x_n=lim s_n+t_n=lim s_n+limt_n=s+t$ and from all the previous steps I would conclude $x \in S+T$. $$$$Maybe I could start by proving that $\{t_n\}_{n \in \mathbb N}$ is convergent. But I don't see why this is true, I've tried to prove that $\{t_n\}_{n \in \mathbb N}$ is a Cauchy sequence (in a Banach space this would imply that the sequence converges) but I couldn't.
One way to do this is as follows : $E/S$ is a Banach space, and the image of $T$, $\pi(T)$, in the quotient is a finite dimensional subspace of $E/S$. Hence, $\pi(T)$ is closed, and so $$ S+T = \pi^{-1}(\pi(T)) $$ is closed in $E$
Added : Here is another approach - by induction on $dim(T)$, we can assume without loss of generality that $dim(T) = 1$. So let $z \in T$ be non-zero, so we are now looking at $$ S+T = \{s + \alpha z : s \in S, \alpha \in \mathbb{K}\} $$ By the Hahn-Banach theorem, there is a continuous linear functional on $E$ such that $$ f(s) = 0 \quad\forall s\in S, \text{ and } f(z) =1 $$ So if $x_n = s_n + \alpha_nz \to x$, then $$ \alpha_n = f(x_n) \to f(x) $$ Hence, $$ s_n = x_n - \alpha_nz\to x-f(x)z $$ Since $S$ is closed, $s:= x-f(x)z\in S$, and hence $$ x = s+f(x)z \in S+T $$