Let $X$ be a normed space, $E:=\{x\in X: \|x\| \leq 1\}.$
The definition of “compact” is the usual one by finite subcover (i.e., not by Dedekind-finite subcover, ultrafilters converge, etc.).
Theorem If $E$ is compact, then $X$ is finite dimensional.
Proof(source)
prove the contrapositive by fixing $\epsilon > 0$ and then choosing a countable set of linearly independent vectors with length less than $1$, so that each subsequent vector is further than $\epsilon$ from the span of the previous vectors. This can be done with Riesz's lemma. This is a sequence with no convergent subsequence.
Question
I think the above proof uses the axiom of dependent choice. Is there a proof of this theorem that uses neither AC nor a weaker version of it?
