Closure of the union = Union of closures

(1) ($\supset$) :: \begin{align*} A \subset A \cup B \implies \text{cl}(A) \subset \text{cl}(A \cup B) \\B \subset A \cup B \implies \text{cl}(B) \subset \text{cl}(A \cup B) \end{align*} therefore yielding that $\text{cl}(A) \cup \text{cl}(B) \subset\text{cl}(A \cup B)$

(2) ($\subset$) ::

The subset $\text{cl}(A) \cup \text{cl}(B)$ is closed and both contains $A$ and $B$, therefore $A \cup B \subset \text{cl}(A) \cup \text{cl}(B)$. $\text{cl}(A \cup B)$ is defined to be smallest closed set which contained $A \cup B$, so that any closed set which contained $A\cup B$ also contains $\text{cl}(A \cup B)$. Therefore $\text{cl}(A \cup B) \subset \text{cl}(A) \cup \text{cl}(B)$.

In your proposed counterexample, you've forgotten that open sets are closed under finite intersection.

So if $x$ has a neighborhood that only meets $A\cup B$ in $B\setminus A$ and a neighborhood that only meets $A\cup B$ in $A\setminus B$, then the intersection of these neighborhoods doesn't meet $A\cup B$ at all.

Let $x\in Cl(A\cup B)$ then every open set containing $x$ intersects $A\cup B$. Thus for $x \in U_\alpha$ where $U_\alpha$ is open in $X$. If $U_\alpha$ intersects $A$, then $x \in Cl(A)$ else $x \in Cl(B)$ either way $x \in Cl(A)\cup Cl(B)$.