In a Banach space X, if $\forall f \in C(M)$, $f$ is bounded, show M is compact

Question

Let $M$ be a set in Banach space $X$ such that $\forall f \in C(M)$, function $f$ is bounded on $M$. Prove that $M$ is compact.

My idea originally was to use Ascoli-Arzela Theorem and then show its closed, but I realize X is not compact, so this makes no sense. Now I'm stuck though and don't even know where to start. We haven't really talked about Banach spaces so the only thing I know is that X being a Banach spaces implies every Cauchy sequence converges to a point in X.

Possible duplicate of Unbounded continuous function on non-compact metric space. See also this similar question: if $f$ has no maximum, then either it is unbounded itself or $g=\frac{1}{-f+\sup f}$ is. — , Feb 24 '18 at 15:48
Are you really asking me to explain why the fact that for all non-compact metric spaces there is an unbounded continuous function implies that all the metric spaces which admit only bounded continuous functions are compact? — , Feb 24 '18 at 15:56

score 2 · Accepted Answer · answered Feb 24 '18 at 15:45

A space $M$ such that all real-valued continuous functions on it are bounded is called pseudocompact. It is well-known that for metrisable spaces (which subsets of Banach spaces are) this property is equivalent to compactness.

(In terms of general topology, first show it's equivalent in normal spaves to limit point compactness; this uses Tietze's extension theorem, and then in first countable spaces this is equivalent again to sequential compactness which also implies Lindelöfness in metric settings, so together implying compactness; there are other routes).

In a Banach space X, if $\forall f \in C(M)$, $f$ is bounded, show M is compact

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