TITLE: Quotient of compact metric space is metrizable (when Hausdorff)?
QUESTION [6 upvotes]: Apparently it's a standard result that, although not every (Hausdorff) quotient of a metric space is metrizable, it always is metrizable when the space being quotiented is compact.
Alas, I can't find a proof - neither in textbooks nor one of my own.
Can anyone give a hint,  or a reference (or a proof, of course)?
Question 2: Is every quotient of a compact metric space (i.e. whether this quotient is Hausdorff or not) second-countable?

REPLY [5 votes]: Here is a somewhat high-tech proof for the first question that I'm fond of.  First, note that if $K$ is a compact Hausdorff space, then $K$ is metrizable iff the space $C(K)$ of continuous real-valued functions on $K$ is separable (with respect to the sup norm).  Indeed, if $K$ is metrizable iff $K$ embeds in the space $[0,1]^\mathbb{N}$, and $K$ embeds in $[0,1]^\mathbb{N}$ iff there are countably many elements of $C(K)$ that separate points of $K$.  If $C(K)$ is separable, you can find countably many such elements by taking a countable dense subset.  Conversely, if countably many functions separate points, then by the Stone-Weierstrass theorem the set of all polynomials in these functions with rational coefficients is dense in $C(K)$, so $C(K)$ is separable.
Now we simply observe that if $X$ is a compact metric space and $K$ is a Hausdorff quotient of $X$, the quotient map $X\to K$ induces an isometric embedding $C(K)\to C(X)$.  Since $X$ is metrizable, $C(X)$ is separable.  Since any subspace of a separable metric space is separable, it follows that $C(K)$ is separable, and hence $K$ is metrizable.