Let $X$ be a topological space, and $A$ a subset of $X$. I want to show that $\bar{A}= A \cup \partial A$.
My proof probably isn't the simplest, but I'm just beginning to write proofs seriously and I want to know if this one holds up. I'll show one inclusion here but the other is almost identical.
Suppose $x\in\bar{A}$ but $x\not\in A \cup \partial A$. Then since $x\not\in\partial A$, there exists a neighbourhood $N$ of $x$ which either does not intersect $A$ or $X\setminus A$. Since $x\in \bar{A}$, every neighbourhood of $x$ intersects $A$, so $N$ must not intersect $X\setminus A$. But this is absurd, since $x\in X\setminus A$.
It's a bit roundabout, as you say, but it looks OK. I'd prove it like this (no proof from contradiction): let $x \in \overline{A}$. If $x \in A$ we're done: $x \in A \cup \partial A$, and if not, every neighbourhood of $x$ intersects $A$ (from being in its closure) and $X\setminus A$ (in $x$ itself at least), so $x \in \partial A$ and we're done too. Same arguments in a more "positive" form.
The reverse is similar: if $x \in A \cup \partial A$, then $x \in A \subseteq \overline{A}$ and we're done or $x \in \partial A \subseteq \overline{A}$ (if every neighbourhood of $x$ intersects both $A$ and its complement, a fortiori every neighbourhood intersects $A$) and we're done again.