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**mabruka** Right, i agree it can not have a countable subcover.

Suppose there is, $\displaystyle O \subset \bigcup_{j=i_1,i_2,...}G_j $ for some numbers $\displaystyle \{i_1,i_2,...\} \subset [0,1] $

Now let $\displaystyle x \in X$ be such an element that $\displaystyle \pi_j (x) = 1 $ for every $\displaystyle j \in \{i_1,i_2,...\}$ where $\displaystyle \pi_j $ is the projection on the jth coordenate.

Clearly $\displaystyle x \not \in\bigcup_{j=i_1,i_2,...}G_j$ but $\displaystyle x \in \bigcup_{j\in[0,1]}G_j = O$

What do you think?