Let $A$ and $B$ be two subsets of a topological space $S$. I would like to prove that $$\partial (A \cap B) \subset \partial(A) \cup \partial(B)$$ where $\partial$ denotes the boundary of that set. I already saw a slick way of proving it on here, but I would like to prove it using cases. Here is my proof.
Let $p \in \partial(A\cap B)$, thus for any open set $U \subset S$ containing $p$ there exists $x, y \in U$ such that $x \in A \cap B$ and $y \not\in A \cap B$. Breaking this down into cases we have: $$x \in A \text{ and } x \in B$$ and one of the following three possibilities: $$\begin{align} y \not\in A &\text{ and } y \not\in B \tag{1}\\ y \in A &\text{ and } y \not\in B \tag{2}\\ y \not\in A &\text{ and } y \in B. \tag{3}\end{align}$$
In case (2), any neighborhood $U$ of $p$ contains a point $x \in B$ and another point $y \not\in B$. Hence $p \in \partial(B)$. Case (3) is identical and shows that $p \in \partial(A)$, and case (1) does both at once. Putting this together we have $$p \in \partial(A) \cup \partial(B).$$
Since $p$ was arbitrary we are done.
Is this correct?