[proofplan]
We intersect each member of the countable base $\mathcal{B}$ for $X$ with $A$ to produce a countable family $\mathcal{B}_A$ of open sets in the subspace topology. The base property of $\mathcal{B}$ transfers directly: every subspace-open set $G \cap A$ (with $G \in \tau$) is a union of members of $\mathcal{B}$, and intersecting each member with $A$ shows that $G \cap A$ is a union of members of $\mathcal{B}_A$.
[/proofplan]
[step:Construct the candidate base $\mathcal{B}_A$ by intersecting basis elements with $A$]
Let $\mathcal{B} = \{B_n\}_{n=1}^\infty$ be a countable base for $(X, \tau)$. Define
\begin{align*}
\mathcal{B}_A := \{B_n \cap A : n \in \mathbb{N}\}.
\end{align*}
Each $B_n \cap A$ is open in the [subspace topology](/page/Subspace%20Topology) on $A$ (by definition, the subspace topology consists of all sets of the form $G \cap A$ with $G \in \tau$, and $B_n \in \tau$). The family $\mathcal{B}_A$ is countable, since it is indexed by $\mathbb{N}$.
[guided]
The subspace topology $\tau_A$ on $A$ consists of all intersections $G \cap A$ where $G \in \tau$. To find a countable base for $\tau_A$, the natural candidate is to take each basis element $B_n$ of the ambient topology and restrict it to $A$.
Let $\mathcal{B} = \{B_n\}_{n=1}^\infty$ be a countable base for $(X, \tau)$. Define
\begin{align*}
\mathcal{B}_A := \{B_n \cap A : n \in \mathbb{N}\}.
\end{align*}
Each member $B_n \cap A$ belongs to $\tau_A$ (it is the intersection of the open set $B_n \in \tau$ with $A$). The family $\mathcal{B}_A$ has at most countably many members (it is indexed by $\mathbb{N}$; some intersections $B_n \cap A$ may coincide or be empty, but the family is still countable). It remains to verify that $\mathcal{B}_A$ is a base for $\tau_A$.
[/guided]
[/step]
[step:Verify that $\mathcal{B}_A$ is a base for the subspace topology on $A$]
Let $V \in \tau_A$. By definition of the subspace topology, $V = G \cap A$ for some $G \in \tau$. Since $\mathcal{B}$ is a base for $\tau$, the open set $G$ is a union of basis elements:
\begin{align*}
G = \bigcup_{n \in I} B_n
\end{align*}
for some index set $I \subset \mathbb{N}$. Intersecting both sides with $A$:
\begin{align*}
V = G \cap A = \left(\bigcup_{n \in I} B_n\right) \cap A = \bigcup_{n \in I} (B_n \cap A).
\end{align*}
Each $B_n \cap A$ is a member of $\mathcal{B}_A$, so $V$ is a union of members of $\mathcal{B}_A$. Since $V \in \tau_A$ was arbitrary, $\mathcal{B}_A$ is a countable base for the subspace topology on $A$, confirming that $A$ is second-countable.
[guided]
We must show that every open set in $\tau_A$ is a union of members of $\mathcal{B}_A$. Let $V \in \tau_A$ be arbitrary. By the definition of the subspace topology, there exists $G \in \tau$ with $V = G \cap A$.
Since $\mathcal{B}$ is a base for $\tau$, we may write $G$ as a union of basis elements:
\begin{align*}
G = \bigcup_{n \in I} B_n
\end{align*}
for some index set $I \subset \mathbb{N}$. Intersecting both sides with $A$ and distributing the intersection over the union (which is valid for arbitrary families of sets):
\begin{align*}
V = G \cap A = \left(\bigcup_{n \in I} B_n\right) \cap A = \bigcup_{n \in I} (B_n \cap A).
\end{align*}
Each set $B_n \cap A$ belongs to $\mathcal{B}_A$ by definition. Therefore $V$ is a union of members of $\mathcal{B}_A$. Since this holds for every $V \in \tau_A$, the family $\mathcal{B}_A$ is a base for the subspace topology.
The family $\mathcal{B}_A$ is countable (indexed by $\mathbb{N}$), so $(A, \tau_A)$ is second-countable. Note that the argument uses no special properties of $A$ --- it applies to every subset, whether open, closed, or neither.
[/guided]
[/step]
