TITLE: Compactness gives Second countable space??
QUESTION [5 upvotes]: Suppose that $(X,\mathcal{T})$ is a topological space. If we know that $X$ is compact can we assume that is also second countable ??
Because X is compact we have that $X=\cup_{i=1}^{n}V_{i}$, where $V_{i}$ are open sets. So for every $U\subset X$ we have that $U\subset \cup_{i=1}^{n}V_{i}$. Can we assume that by choosing some of the $V_{i}$ we will have that $U=\cup_{i\in J}V_{i}$, $J$ is countable.
In order for $(X,\mathcal{T})$ to have a countable topology base ??

REPLY [2 votes]: The $\epsilon$-order  topology on the ordinal $\omega_1+1$  is compact Hausdorff but not 2nd-countable. It has an uncountable subset of isolated points: $\{x+1:x\in \omega_1\}\cup \{0\}.$
It is not even 1st-countable: The  point $ \omega_1$ has uncountable character.