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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 "meta-proof" (i do not construct $\epsilon$-nets in details).

Let $X = [0,1]\times[0,\frac{1}{2}]\times[0,\frac{1}{3}]\times\dotsb$ be a Hilbert cube endowed with its $\ell_2$ metric. For every $k$, let $N_k$ be a "well chosen" $\frac{1}{2^k}$-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 $\frac{1}{2^m}+\frac{1}{2^n}$.

Take "the first" $x_1\in S_1$, 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 $\lbrace x_k \rbrace$ is Cauchy. Its limit is not in any element of $\mathcal U$.

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 $\mathcal U$.

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 "meta-proof" (i do not construct $\epsilon$-nets in details).

Let $X = [0,1]\times[0,\frac{1}{2}]\times[0,\frac{1}{3}]\times\dotsb$ be a Hilbert cube endowed with its $\ell_2$ metric. For every $k$, let $N_k$ be a "well chosen" $\frac{1}{2^k}$-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 compliment 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 $\frac{1}{2^m}+\frac{1}{2^n}$.

Take "the first" $x_1\in S_1$, 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 $\lbrace x_k \rbrace$ is Cauchy. Its limit is not in any element of $\mathcal U$.

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 "meta-proof" (i do not construct $\epsilon$-nets in details).

Let $X = [0,1]\times[0,\frac{1}{2}]\times[0,\frac{1}{3}]\times\dotsb$ be a Hilbert cube endowed with its $\ell_2$ metric. For every $k$, let $N_k$ be a "well chosen" $\frac{1}{2^k}$-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 $\frac{1}{2^m}+\frac{1}{2^n}$.

Take "the first" $x_1\in S_1$, 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 $\lbrace x_k \rbrace$ is Cauchy. Its limit is not in any element of $\mathcal U$.

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 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 $\mathcal U$.

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 "meta-proof" (i do not construct $\epsilon$-nets in details).

Let $X = [0,1]\times[0,\frac{1}{2}]\times[0,\frac{1}{3}]\times\dotsb$ be a Hilbert cube endowed with its $\ell_2$ metric. For every $k$, let $N_k$ be a "well chosen" $\frac{1}{2^k}$-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 compliment 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 $\frac{1}{2^m}+\frac{1}{2^n}$.

Take "the first" $x_1\in S_1$, 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 $\lbrace x_k \rbrace$ is Cauchy. Its limit is not in any element of $\mathcal U$.