$f: [a,b] \to \ell^2$ is a continuous function where $a,b \in \Bbb R$. Then $f([a,b])$ is a compact subspace of $l^2$?

Question

$$\ell^2 =\{\chi | \chi = (x_n) \text{ which is a sequence of real numbers such that} \sum_{n=1}^\infty x_n^2 < \infty\}$$

Original question asks whether it is a proper closed subspace.

What I found so far: In a different question an answer by Henno Brandsma states that $f([a,b])$ is a compact subspace and in a metric space it automatically implies it is closed, which is correct. Also, $\ell^2$ is not compact (I can prove this by taking the basis vectors $e_k$ which are bounded but do not converge in a norm in any fashion). Therefore it is a a proper closed subspace.

My question is thus how to prove that $f([a,b])$ is a compact subspace; which would take care of the entire problem, as the rest is done.

Answer 1

That happens because continuous maps map compact sets into compact sets. And $[a,b]$ is compact.