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?
Please help me to know it.

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.
