Do you know if the following statement is an equivalent form of the axiom of choice or not?



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> If $X$ is a compact metric space, then every continuous function $f: X \longrightarrow \mathbb{R}$ is uniformly continuous.
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If you know any references, please let me know.



Axiom of Choice and continuous functions
It seems to me that this is provable without using the axiom of
choice.


Suppose that $X$ is a compact metric space and $f:X\to\mathbb{R}$
is continuous. Let's show it is uniformly continuous. Fix any
$\epsilon\gt 0$. For each point $x\in X$, there is a small ball
$B$ centered at $x$ such that $f(y)$ is within $\epsilon/2$ of
$f(x)$ for all $y\in B$, and we may choose $B$ to have radius
$1/{n\_x}$, where we choose $n\_x$ to the smallest positive natural number for which this radius has the desired property. Thus, we have a canonical choice of radius here, and so we
do not need the axiom of choice to define the map $x\mapsto n\_x$.


Consider now the family $\{ B\_{1/{2n\_x}}(x) \mid x\in X \}$, consisting of the inner core of each of those balls, with the radius of each of them shrunk to half. This is an open cover of $X$, and so by
compactness there is a finite subcover, which consists of finitely
many balls, having some minimal radius $1/{2n}$. Now, if $y$ and
$z$ are within $1/{2n}$ of each other, then they are both within
$1/{n\_x}$ of the center $x$ of the ball in which $y$ sits in the
subcover, and so $f(y)$ and $f(z)$ are within $\epsilon/2$ of
$f(x)$, and hence within $\epsilon$ of each other, showing that
$f$ is uniformly continuous.


