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.
Update. Here is a proof.
Let $X = [-\frac{1}{2},\frac{1}{2}]\times[-\frac{1}{4},\frac{1}{4}]\times[-\frac{1}{8},\frac{1}{8}]\times\dotsb$ be a Hilbert cube endowed with its $\ell_\infty$ metric. For every positive integer $k$, let $N_k$ be the “natural” $\frac{1}{2^k}$-net for $X$.
Let $\mathcal U$ be a given family of open sets such that no finite subfamily of $\mathcal U$ covers $X$. Then let $S_k$ be the set of those elements of $N_k$ which are within the distance of $\frac{1}{2^k}$ from the complement of any finite union of elements of $\mathcal U$. Each $S_k$ is nonempty. For all $m$ and $n$, the distance from any point of $S_m$ to the set $S_n$ is at most $\frac{1}{2^m}+\frac{1}{2^n}$.
Assume that the points of $X$ are ordered by the lexicographic order of their coordinates. Take the “first” $x_1\in S_1$ (i.e. the smallest in the order), then the “first” $x_2\in S_2$ that is within the distance of $\frac{3}{4}$ from $x_1$, then the “first” $x_3\in S_3$ that is within the distance of $\frac{3}{8}$ from $x_2$, and so forth. The obtained sequence $\lbrace x_k \rbrace_{k=1}^\infty$ is Cauchy. Its limit is not in any element of $\mathcal U$.