I'm trying to prove that $\overline{A\cup B}=\overline{A}\cup\overline{B}$, with $A,B\subset X$ and closure defined by $$\overline{A}=\{x\in X:\text{every neighbourhood of $x$ intersects A}\}$$
I first show that $\overline{A\cup B}\subset\overline{A}\cup\overline{B}$, so let $x\in\overline{A\cup B}$
Then every neighbourhood $U$ of $x$ intersects ${A\cup B}$, i.e $$(A\cup B)\cap U\neq \emptyset\implies (A\cap U)\cup(B\cap U)\neq\emptyset$$ which means that every neighbourhood of $x$ intersects $A$, or $B$, or both, i.e $x\in\overline{A}\cup\overline{B}$. Is this correct?
For the other direction could I start with $(A\cap U)\cup(B\cap U)\neq\emptyset$ and just reverse the steps?
No, in your setup, every neighbourhood $U$ intersects $A$ or $B$. But it has to be always $A$ to conclude that $x \in \overline{A}$ and likewise for $B$. But you only know that you intersect one of them for a given $U$.
That's why it's better to go by contradiction: suppose $x \in \overline{A \cup B}$ but $x \notin \overline{A} \cup \overline{B}$, so $x \notin \overline{A}$ and $x \notin \overline{B}$. Not being in the closure of $A$ gives you that there is a neighbourhood of $x$, say $U_1$ such that $U_1 \cap A = \emptyset$, and not being in $\overline{B}$ another neighbourhood $U_2$ of $x$ such that $B \cap U_2 = \emptyset$. Now...
For the other direction: $A \subseteq A \cup B$ so $\overline{A} \subseteq \overline{A \cup B}$, and also for $B$, hence....