Why not indulge in some overkill:
Theorem (cf. Federer, Theorem 4.5.11). $\newcommand{\RR}{\mathbb{R}}A\subset\RR^n$ has finite perimeter iff $\mathcal{H}^{n-1}(\partial^* A)<\infty$.
Here $\mathcal{H}^s$ denotes the $s$-dimensional Hausdorff measure and $\partial^*A\subset\RR^n$ the measure-theoretic boundary of $A$. The latter is determined by $$ x\in\RR^n\setminus\partial^*A \quad\Longleftrightarrow\quad \lim_{r\to 0+}\frac{\mathcal{H}^n(A\cap B(x,r))}{\mathcal{H}^n(B(x,r))}\in\{0,1\}. $$ In other words, $x\in\partial^*A$ iff $A$ does not have Lebesgue density $0$ or $1$ at $x$.
By Federer's theorem it suffices to show that $\partial^*(S\cap T)\subset\partial^*S\cup\partial^*T$, or equivalently, $$(\RR^n\setminus\partial^*S)\cap(\RR^n\setminus\partial^*T)\subset\RR^n\setminus\partial^*(S\cap T).$$
The rest is elementary. Let $x$ be an element of the left hand side. Clearly, if $S$ or $T$ has density $0$ at $x$, then $S\cap T$ has density $0$ at $x$. So suppose both $S$ and $T$ have density $1$ at $x$. But as $$(S\cap B(x,r))\setminus (B(x,r)\setminus T) = S\cap T\cap B(x,r),$$ also $S\cap T$ has density $1$ at $z$. So $x$ is an element of the right hand side.