Prove for all convergent sequence $\phi\rightarrow x_0$, the set $A=\{x_0\}\cup \phi(\mathbb{N})$ is compact
My work
Let $U=\{\bigcup_\alpha U_\alpha:\alpha\in\sum\}$ a open covers of $A$, then for some $U_\alpha$, we have $x_0\in U_\alpha$. Then for $\epsilon>0$ we have $B(x_0,\epsilon)\subset U_0$.
Here i'm little stuck, because my space $(X,\tau)$ not necessarily is a metric space. I'm a little confused here. Can someone help me?
Recall the definition of convergence in a general topological space: $L$ is a limit of a sequence if for every neighborhood of $L$ the sequence is eventually in that neighborhood. It doesn't depend on balls or distances.
So once you have chosen an $U_\alpha$ that contains $x_0$, it will also contain all but finitely many elements of the sequence. Finitely many additional $U_\alpha$s will cover the rest.