Hilbert Cube is Compact

Theorem
Let $M = \left({I^\omega, d_2}\right)$ be the Hilbert cube.

Then $M$ is a compact space.

Proof
Let $M'$ be the metric space defined as:
 * $M' = \displaystyle \prod_{k \mathop \in \N} \left[{0 \,.\,.\, 1}\right]$

under the Tychonoff topology.

By definition, $\left[{0 \,.\,.\, 1}\right]$ is the closed unit interval under the usual (Euclidean) topology.

From Hilbert Cube is Homeomorphic to Countable Infinite Product of Real Number Unit Intervals, $M$ is homeomorphic to $M'$.

From Closed Real Interval is Compact, $\left[{0 \,.\,.\, 1}\right]$ is a compact space.

But $M'$ is also a compact space by Tychonoff's Theorem.

From Compactness is Preserved under Continuous Surjection, if $M'$ is compact then so is $M$.