The space $X$ is called limit point compact if every infinite subset $E$ of $X$ has a limit point. It can also be proved that (open cover) compactness implies limit point compactness.
Let $[0,1]^\omega$ have the product topology. Clearly $[0,1]^\omega$ is compact, so therefore it should be limit point compact. But the infinite subset $\{(1,0,0,0,...),(0,1,0,0,...),(0,0,1,0,...),(0,0,0,1,...),...\}$ doesn't seem to have a limit point.
I have a suspicion that $(0,0,0,0,...)$ is a limit point of this set. But how do you show this formally? We can easily find an open set around $(0,0,0,0,...)$ which does not contain any sequence with a $1$ in it.
Take a neighborhood of $(0,0,...)$. Inside that neighborhood there is an open set $U$ of the form $U_1\times U_2\times ...\times U_N\times [0,1]^\omega$, where $U_i$ are open neighborhoods of $0$ in $[0,1]$.
Then all elements of the sequence that have index larger than $N$ are in $U$.
Your last sentence contains a misunderstanding of the definition of the product topology. I suspect that you are thinking that neighborhoods of $(0,0,...)$ can be for the form $U_1\times U_2\times U_3,...$ where the $U_i$ are arbitrary open neighborhoods of $0$ in $[0,1]$. However, note that in the product topology only finitely many of those $U_i$ can be different from $[0,1]$.