The compactness of the Hilbert cube follows (without AC) from the compactness of $2^\omega$, since $[0,1]$ as well as $[0,1]^\omega$ are continuous images of $2^\omega$. 

(Conversely, $2^\omega$ is a closed subset of the Hilbert cube.) 

The compactness of $2^\omega$ is just König's lemma for trees of binary sequences, which is easy to prove (hence certainly well-known) without AC.