Each subspace is compact, Hausdorff and metrizable. Therefore, second-countable (Exercise 3). According to the same exercise, we only need to show that $X$ is second-countable. If $X_1$ and $X_2$ were disjoint, they would be open as well, and we could take the union of their bases: if $x\in X_i$ then every its neighborhood $U$ in $X$ has a sub-neighborhood $U\cap X_i$ , open in $X_i$ and $X$ , and, therefore, it would contain a basis neighborhood open in $X_i$ and $X$ . If they are not disjoint, then we do not even know that a basis set in $X_i$ is open in $X$ . Let $\{U^i_n\}$ be a basis in $X_i$ . Suppose $U$ is a neighborhood of $x\in X$ . If $x\in X_i-X_j$ then $U\cap X_i$ is open in $X_i$ and there is a basis neighborhood $U^i_n$ containing $x$ . Moreover, $x\in U^i_n\cap X-X_j$ is open in $X-X_j$ which is open in $X$ , therefore, $U^i_n\cap X-X_j$ is open in $X$ as well. This suggests that, first, we take all intersections $\{U^i_n\cap X_i-X_j\}$ (they are all open in $X$ ). Note that if $X_1$ and $X_2$ were disjoint this would be exactly the union of their bases. In either case, this already provides bases for all points not in $X_1\cap X_2$ . Now, suppose $x\in X_1\cap X_2$ . $U\cap X_i$ has a basis sub-neighborhood $U^i_{n_i}$ of $x$ in $X_i$ . $U^i_{n_i}=X_i\cap V^i_{n_i}$ where $V^i_{n_i}$ is open in $X$ . $x\in V^1_{n_1}\cap V^2_{n_2}=$ $(V^1_{n_1}\cap V^2_{n_2}\cap X_1)\cup$ $(V^1_{n_1}\cap V^2_{n_2}\cap X_2)=$ $(U^1_{n_1}\cap V^2_{n_2})\cup$ $(V^1_{n_1}\cap U^2_{n_2})$ $\subseteq U$ . So, for every pair of intersecting basis sets in $X_1$ and $X_2$ choose some open sets in $X$ that intersect with the subspaces at these basis sets and take their intersection. We have a countable basis for points in $X_1\cap X_2$ .