Thanks for all the answers and sorry about a silly question. I have also figured out that it can be proved using the usual complete metric on the usual (countable product) Hilbert cube and finite $\epsilon$-nets.

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**Update.** Here is my attempt to fix this proof, please tell me if i have missed something. This is a "meta-proof" (i do not construct $\epsilon$-nets in details).

For every $k$, let $N_k$ be a "well chosen" $\frac{1}{2^k}$-net for $X = [0,1]^\omega$. Suppose $\mathcal U$ is an open cover of $X$ without a finite subcover. Let $S_k$ be the set of those elements of $N_k$ which are within the distance of $\frac{1}{2^k}$ from the compliment of any finite intersection of elements of $\mathcal U$. Each $S_k$ is nonempty. For all $m$ and $n$, the distance between any point of $S_m$ and the set $S_n$ is at most $\frac{1}{2^m}+\frac{1}{2^n}$.

Take "the first" $x_1\in S_1$, then "the first" $x_2\in S_2$ within the distance of $\frac{3}{4}$ from $x_1$, then "the first" $x_3\in S_3$ within the distance of $\frac{3}{8}$ from $x_2$, etc. The obtained sequence $\lbrace x_k \rbrace$ is Cauchy, and its limit is not in any element of $\mathcal U$.