# Is every compact metric space hereditarily separable?

Let $X$ be a compact metric space. I see why all open and closed subsets of $X$ are separable. But is every subset of $X$ necessarily separable?

EDIT: Since $X$ is separable metric, it embeds into the Hilbert cube $[0,1]^\omega$, which is hereditarily separable, right? And so $X$ is also hereditarily separable.

• If I recall correctly, every separable metric space is second countable. That should do it. – Pedro Sánchez Terraf Feb 11 '16 at 3:18
• Countable Choice turns out to be necessary and sufficient. See here and here. – Cameron Buie Feb 11 '16 at 3:37