Hilbert Cube is Separable

Theorem
Let $M = \struct {I^\omega, d_2}$ be the Hilbert cube.

Then $M$ is a separable space.

Proof
Consider the set $H$ of all points of $M$ which have finitely many rational coordinates and all the rest zero.

Then $H$ forms a countable subset of $A$ which is (everywhere) dense.

The result follows by definition of separable space.