Is a compact metric space with a "doubled" point sequentially compact?

Let $$X$$ be a compact metric space with topology $$\tau$$ generated by the metric. Consider a new point $$x_1\notin X$$ and a non-isolated point $$x_0\in X$$. Set $$\overline{X}=X\cup \{x_1\}$$, equipped with the topology

$$\begin{equation} \overline{\tau}= \tau \cup \left\{W\cup \{x_1\}: x_0\in W, W\in\tau\right\}\cup \left\{(W\setminus \{x_0\})\cup \{x_1\}: x_0\in W, W\in \tau\right\}. \end{equation}$$ I think that $$(\overline{X}, \overline{\tau})$$ is a compact space. Also It is not metrizable, because $$\overline{X}$$ is not a Hausdorff space.

Can we say that $$\overline{X}$$ is sequentially compact?

• @evaristegd, Thanks for your reply. I would like to know that $\overline{X}$ is a sequentially compact? Jul 17 '19 at 4:59
Yes, $$\overline{X}$$ is sequentially compact. Given a sequence $$(a_n)$$ in $$\overline{X}$$, if $$a_n\in X$$ for infinitely many $$n$$, then we can find a subsequence which converges in $$X$$ and hence in $$\overline{X}$$. The only other possibility is that $$a_n=x_0$$ for all but finitely many $$n$$, and then $$(a_n)$$ converges to $$x_0$$.
More generally, any space which is a finite union of sequentially compact subspaces is sequentially compact (here the subspaces are $$X$$ and $$\{x_0\}$$). The proof is similar: given a sequence, it must have infinitely many terms in one of the subspaces, and so it has a convergent subsequence in that subspace.