Your proof for (2) is basically fine, though the claim ‘Then $U \times V = {(x,x)}$’ could use a little more explicit justfication.
For (3), let $x \in \operatorname{bdry}(A \cup B)$; then $x \in \operatorname{cl}(A \cup B) = \operatorname{cl}A \cup \operatorname{cl}B$ and $x \in \operatorname{cl}\left(X \setminus (A \cup B)\right)$. Without loss of generality say $x \in \operatorname{cl}A$; I claim that $x \in \operatorname{bdry}A$. If not, then $x \notin \operatorname{cl}(X \setminus A)$, and there is an open set $V$ such that $x \in V$ and $V \cap (X \setminus A) = \varnothing$. But then $x \in V \subseteq A \subseteq A \cup B$, so $V$ is an open nbhd of $x$ that doesn’t intersect $X \setminus (A \cup B)$, and $x \notin \operatorname{cl}\left(X \setminus (A \cup B)\right)$. This contradiction establishes the claim, and it follows that $x \in \operatorname{bdry}(X \setminus A) \subseteq \operatorname{bdry}A \cup \operatorname{bdry}B$ and hence that $\operatorname{bdry}(A \cup B) \subseteq \operatorname{bdry}A \cup \operatorname{bdry}B$.
Now let $x \in \operatorname{bdry}A \cup \operatorname{bdry}B$; we may assume that $x \in \operatorname{bdry}A$, so that $x \in \operatorname{cl}A$ and $x \in \operatorname{cl}(X \setminus A)$. Certainly we then have $x \in \operatorname{cl}A \subseteq \operatorname{cl}(A \cup B)$, so to show that $x \in \operatorname{bdry}(A \cup B)$, we need only show that $x \in \operatorname{cl}\left(X \setminus (A \cup B)\right)$. By hypothesis $A \cap \operatorname{cl}B = B \cap \operatorname{cl}A = \varnothing$, so $\operatorname{cl}B \subseteq X \setminus A$ and $\operatorname{cl}A \subseteq X \setminus B$. But then $\operatorname{cl}\left(X \setminus (A \cup B)\right) = \operatorname{cl}\left((X \setminus A) \cap (X \setminus B) \right) \supseteq \operatorname{cl}\left(\operatorname{cl}B \cap \operatorname{cl}A \right) =$ $\operatorname{cl}B \cap \operatorname{cl}A$. Certainly $x \in \operatorname{cl}A$, so we’re done if $x \in \operatorname{cl}B$.
Edit:
If $x \notin \operatorname{cl}B$, let $V = X \setminus \operatorname{cl}B$, so that $V$ is an open nbhd of $x$ contained in $X \setminus B$. Let $W$ be any open nbhd of $x$; then $\begin{align*}W \cap \left(X \setminus (A \cup B) \right) &= W \cap (X \setminus A) \cap (X \setminus B)\\ &\supseteq W \cap (X \setminus A) \cap V\\ &= (V \cap W) \cap (X \setminus A)\\ &\ne \varnothing, \end{align*}$ since $x \in \operatorname{cl}(X \setminus A)$, and hence $x \in \operatorname{cl}\left(X \setminus(A \cup B)\right)$ anyway.
To see what’s really going on, notice that in principle $\operatorname{bdry}(A \cup B)$ can fail to contain $\operatorname{bdry}A$. This situation can already be illustrated in the real line, with $A = [0,1)$ and $B = [1,2]$, for instance. Here $1 \in \operatorname{bdry}A = \{0,1\}$, but $1 \notin \operatorname{bdry}(A \cup B) = \{0,2\}$. The hypothesis that $A \cap \operatorname{cl}B = B \cap \operatorname{cl}A = \varnothing$ rules out this kind of situation by ensuring that a point in $\operatorname{bdry}A$ cannot already belong to $B$ and so cannot end up in $\operatorname{int}(A \cup B)$ (and similarly with $A$ and $B$ interchanged).
(There are various ways to define the boundary of a set $A$; here I’ve used the definition $\operatorname{bdry}A = \operatorname{cl}A \cap \operatorname{cl}(X \setminus A)$.)