I have found this paper by Peter Loeb:

• Peter A. Loeb, A new proof of the Tychonoff Theorem, The American Mathematical Monthly 72 (1965), no. 7, 711--717.

Here is the theorem from this paper that implies that the usual Hilbert cube is compact without using the AC.

Theorem 1. Let $\{\,X_\nu\mid\nu\in I\,\}$ be a family of compact spaces which is indexed by a set $I$ on which there is a well-ordering $\ge$. If $I$ is an infinite set, let there also be a choice function $F$ on the collection $\{\,C\mid\text{$C$ is closed},\ C\ne\varnothing,\ \text{$C\subset X_\nu$ for some $\nu$}\,\}$. Then the product space $\prod_{\nu\in I}X_\nu$ is compact in the product topology.

I have found this paper by Peter Loeb:

• Peter A. Loeb, A new proof of the Tychonoff Theorem, The American Mathematical Monthly 72 (1965), no. 7, 711--717.

Here is the theorem from this paper that implies that the usual Hilbert cube is compact without using the AC.

Theorem 1. Let $\{\,X_\nu\mid\nu\in I\,\}$ be a family of compact spaces which is indexed by a set $I$ on which there is a well-ordering $\ge$. If $I$ is an infinite set, let there also be a choice function $F$ on the collection $\{\,C\mid\text{$C$ is closed},\ C\ne\varnothing,\ \text{$C\subset X_\nu$ for some $\nu$}\,\}$. Then the product space $\prod_{\nu\in I}X_\nu$ is compact in the product topology.

For the usual Hilbert cube $[0,1]^{\mathbb N}$, the function $F$ can, for example, select the least element in every compact subset of $[0,1]$.