Riesz's Theorem of compactness

Question

$\left(X,\|\cdot\|\right)$ is a normed vector space. $\textbf{Riesz's Theorem of compactness}$ says that $$ \{x \in X \colon \|x\| \leq 1 \} \ \text{compact} \ \Longleftrightarrow \ \text{Each bounded sequence in $X$ has a convergent subsequence}.$$

I am looking for the proof of this theorem.

Some thoughts to $"\Rightarrow"$: If $\{x \in X \colon \|x\| \leq 1 \}$ is compact, it's clear that each sequence in $\{x \in X \colon \|x\| \leq 1 \}$ has a convergent subsequence. But how can I conclude that each bounded sequence in $X$ (!?) has a convergent subsequence?

Answer 1

A bounded set is contained in $B_r(0) = \{x \in X : \|x\| \leq r\}$. The map that sends $x\to rx$ is a homeomorphism of $X$, so if $B_1(0)$ is compact, so is $B_r(0)$. Hence, your bounded sequence is contained in a compact set, and so has a convergent subsequence.

Answer 2

HINT: If $x_n$ is a bounded sequence, and $M$ is a bound on $\|x_n\|$, consider the sequence $\frac1Mx_n$.