$S=\operatorname{cl}_S(A), T = \operatorname{cl}_T(B) \implies \operatorname{cl}_{S\cup T}(A \cup B) = S \cup T$

Question

Let $X$ be a topological space with subsets $A \subseteq S \subseteq X$ and $B \subseteq T \subseteq X$ such that $S=\operatorname{cl}_S(A), T = \operatorname{cl}_T(B)$. Prove that $\operatorname{cl}_{S\cup T}(A \cup B) = S \cup T$.

Note: all subsets have subspace topologies.

Attempt: Let $x \in S \cup T$. WLOG, $x \in S$. Thus since $S = \operatorname{cl}_S(A)$, we know that $V \cap A \neq \emptyset$ for all (open) neighborhoods $V$ of $x$ in $S$.

Suppose to the contrary there is a neighborhood $W$ of $x$ in $S \cup T$ with $W \cap (A \cup B) = \emptyset$. Then, by possibly shrinking $W$, we may assume that $W$ is open in $S \cup T$. This means that $W = (S \cup T) \cap G = (S \cap G) \cup(T \cap G)$ for some subset $G$ open in $X$. Now, also $W \cap A = \emptyset$ and thus $(S \cap G) \cap A= \emptyset$. But $S \cap G$ is open in $S$ and contains $x$. Thus $S \cap G$ is a neighborhood of $x$ in $S$, contradicting what we wrote at the beginning of the proof.

Thus, every neighborhood $W$ of $x$ in $S \cup T$ has $W \cap (A \cup B) \neq \emptyset$ and it follows that $x \in \operatorname{cl}_{S \cup T}(A \cap B)$.

Is this correct?

Answer 1

Direct Proof.
A is dense within S and B is dense within T
implies A $\cup$ B is dense within S $\cup$ T.
Proof.
Assume x in S $\cup$ T.
Let V be open within S $\cup$ T.
Exists X-open U with V = U $\cap$ (S $\cup$ T).
Case x in S: x in S-open U $\cap$ S.
. . As A is dense with in S, exists a in A $\cap$ U $\cap$ S.
Case x in T: Likewise exists b in B $\cap$ U $\cap$ T.
As every not empty (S $\cup$ T)-open set intersects A $\cup$ B,
A $\cup$ B is dense within S $\cup$ T.

Answer 2

For a subspace $S$ we have a formula for the closure in a subspace: if $A \subseteq S \subseteq X$ then $$\operatorname{cl}_S(A)=\operatorname{cl}(A) \cap S$$

where the unmarked closure is taken in the whole space $X$.

So $$\operatorname{cl}_{S \cup T}(A \cup B)=(S \cup T) \cap \operatorname{cl}(A \cup B) = (S \cup T) \cap (\operatorname{cl}(A) \cup \operatorname{cl}(B))$$

and the right hand side already contains the sets (it's a union of 4 intersections, two of which are:) $\operatorname{cl}(A) \cap S = \operatorname{cl}_S(A)=S$ and $\operatorname{cl}(B) \cap T = \operatorname{cl}_T(B)=T$ and so

$$\operatorname{cl}_{S \cup T}(A \cup B)=S \cup T$$

follows trivially.