[proofplan]
The argument routes through second countability. Since $X$ is separable and metrizable, the [Equivalence of Separability Conditions in Metrizable Spaces](/theorems/545) provides a [countable](/page/Countable%20Set) base $\mathcal{B}$ for the [topology](/page/Topology) of $X$. We restrict $\mathcal{B}$ to the subspace $A$ by intersecting each member with $A$, producing a countable base for the subspace topology on $A$. Since $A$ inherits a metric from $X$ and is now second-countable, the same equivalence yields a countable [dense subset](/page/Dense%20Subset) of $A$.
[/proofplan]
[step:Lift separability of $X$ to second countability via the metrizable equivalence]
Let $(X, d)$ be a separable metrizable space and let $A \subset X$ be an arbitrary subspace, equipped with the restriction $d|_{A \times A}$ and the subspace [topology](/page/Topology). Since $X$ is both separable and metrizable, the [Equivalence of Separability Conditions in Metrizable Spaces](/theorems/545) (implication $(1) \Rightarrow (2)$) provides a [countable](/page/Countable%20Set) base $\mathcal{B} = \{B_1, B_2, \ldots\}$ for the topology of $X$.
[/step]
[step:Restrict the countable base to the subspace $A$ to obtain a countable base for the subspace topology]
Define the family
\begin{align*}
\mathcal{B}_A := \{B_k \cap A : k \in \mathbb{N}\}.
\end{align*}
This collection is countable (it is the image of the [countable set](/page/Countable%20Set) $\mathcal{B}$ under the map $B_k \mapsto B_k \cap A$).
[claim:The restricted family $\mathcal{B}_A$ is a base for the subspace topology on $A$]
Every [open set](/page/Open%20Set) in the subspace topology on $A$ is a union of members of $\mathcal{B}_A$.
[/claim]
[proof]
Let $G_A \subset A$ be open in the subspace topology. By the definition of the subspace topology, there exists an open set $G \subset X$ with $G_A = G \cap A$. Since $\mathcal{B}$ is a base for the topology of $X$, the set $G$ can be written as a union of basis elements: $G = \bigcup_{k \in S} B_k$ for some index set $S \subset \mathbb{N}$. Intersecting with $A$:
\begin{align*}
G_A = G \cap A = \left(\bigcup_{k \in S} B_k\right) \cap A = \bigcup_{k \in S} (B_k \cap A).
\end{align*}
Each $B_k \cap A$ belongs to $\mathcal{B}_A$, so $G_A$ is a union of members of $\mathcal{B}_A$.
[/proof]
[guided]
The subspace topology on $A$ consists of all sets of the form $G \cap A$ where $G$ is open in $X$. We need to verify that every such set can be expressed as a union of members of the restricted family $\mathcal{B}_A = \{B_k \cap A : k \in \mathbb{N}\}$.
Why does this work? The key point is that writing an open set $G \subset X$ as a union of basis elements is a purely set-theoretic operation that commutes with intersection. Specifically: if $G = \bigcup_{k \in S} B_k$, then the general distributive law for set intersection gives
\begin{align*}
G \cap A = \left(\bigcup_{k \in S} B_k\right) \cap A = \bigcup_{k \in S} (B_k \cap A).
\end{align*}
This identity converts a representation of $G$ in terms of $\mathcal{B}$ into a representation of $G \cap A$ in terms of $\mathcal{B}_A$. The index set $S \subset \mathbb{N}$ is the same on both sides — we do not need $S$ to be finite or to make any additional choices. Since every member $B_k \cap A$ of $\mathcal{B}_A$ is open in the subspace topology (it is the intersection of the open set $B_k$ with $A$), the family $\mathcal{B}_A$ is indeed a base for the subspace topology on $A$.
Note that this argument uses no metric structure — the restriction of a countable base to a subspace produces a countable base in any topological space. The metric enters only through the prerequisite (theorem 545) that converts separability to second countability.
[/guided]
[/step]
[step:Conclude separability of $A$ from second countability via the metrizable equivalence]
The subspace $A$, equipped with the restricted metric $d|_{A \times A}$, is a metrizable [topological](/page/Topology) space with the [countable](/page/Countable%20Set) base $\mathcal{B}_A$. Applying the [Equivalence of Separability Conditions in Metrizable Spaces](/theorems/545) (implication $(2) \Rightarrow (1)$) to $A$, the space $A$ is separable.
Since $A \subset X$ was an arbitrary subspace, every subspace of $X$ is separable.
[/step]
