TITLE: Axiom of Choice and continuous functions
QUESTION [8 upvotes]: Do you know if the following statement is an equivalent form of the axiom of choice or not?

If $X$ is a compact metric space, then every continuous function $f: X \longrightarrow \mathbb{R}$ is uniformly continuous.

If you know any references, please let me know.

REPLY [2 votes]: Look Ma, no axiom of choice!
THEOREM 0   Let   $X$   be a compact space. Let   $\Phi$   be a non-empty family of closed subsets of   $X$,   $F := \bigcap \Phi$,   and   $G\supseteq F$   an open subset of   $X$.   Then there exists a finite   $\Phi_0\subseteq \Phi$   such that   $\bigcap\Phi_0\subseteq G$.
PROOF   Family   $\Gamma\ :=\ G\cup\{X\setminus A : A \in \Phi\}$   is an open covering of   $X$. Etc.   END of PROOF

As an instant corollary we get:
THEOREM 1   Let   $X$   be a compact space. Let   $\Phi$   be a non-empty family of closed subsets of   $X$,   $F := \bigcap \Phi$,   and   $G\supseteq F$   an open subset of   $X$.   Assume also that   $\Phi$   is linearly ordered by   $\subseteq$.   Then there exists   $A\in \Phi$   such that   $A \subseteq G$.
THEOREM 2   Let   $(X\ d)$   be a compact metric space. Let   $W$   be an open subset of   $X^2$,   such that   $\Delta_X\subset W$,   where   $\Delta_X := \{(x\ x):x\in X\}$.   Then there exists   $\delta > 0$   such that   $V_X(\delta)\subseteq W$,   where   $V_X(\delta) := \{(x\ y)\in X^2 : d(x\ y)\le\delta\}$.
PROOF   Apply Theorem 1 to   $X^2$ as a replacement of   $X$   of Theorem 1; etc.   END of PROOF
THEOREM 3   Let   $f : X\rightarrow Y$   be an arbitrary continuous function of a metric compact space   $(X\ d_X)$   into an arbitrary metric space   $(Y\ d_Y)$.   Then function   $f$   is uniformly continuous.
PROOF   Let   $\epsilon > 0$.   Let   $W\subseteq X^2$   be the inverse image of   $V_Y(\epsilon)$   under function   $f\times f$.   There exists, by Theorem 2,   $\delta > 0$   such that   $V_X(\delta)\subseteq W$.   Then   $d_Y(f(x')\ f(x''))\le\epsilon$   for every   $x'\ x''\in X$   such that   $d_X(x'\ x'')\le\delta$.   END of PROOF