That is true. Caccioppoli sets are also known as sets of finite perimeter.
Theorem. Suppose $f\in L^1(\mathbb{R}^n)$ vanishes outside the unit cube $[0,1]^n$. For $i=1,2,\ldots,n$ consider the function $V_if(x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n)= V_0^1f(x_1,\ldots,x_{i-1},\cdot,x_{i+1},\ldots,x_n)$, i.e. the (essential) variation of the one dimensional sections. Then $f\in BV(\mathbb{R}^n)$ if and only if for every $i$, $V_i\in L^1([0,1]^{n-1}$).
This is Theorem 5.3.5 in W. P. Ziemer, Weakly differentiable functions. Sobolev spaces and functions of bounded variation. Graduate Texts in Mathematics, 120. Springer-Verlag, New York, 1989. Basically it is a characterization of the functions of bounded variation by one dimensional slices.
If $h_1,h_2$ are characteristic functions of sets in $[0,1]$, then $V_0^1(h_1h_2)\leq V_0^1h_1+V_0^1 h_2$, see the proof of Theorem 2 in http://mathonline.wikidot.com/multiples-and-products-of-functions-of-bounded-variation. Now if $S$ and $T$ are Caccioppoli sets cotained in the unit cube, the characteristic functions $\chi_S$ and $\chi_T$ have bounded variation and the one dimensional result mentioned here shows that $$ V_i(\chi_S\chi_T)\leq V_i(\chi_S)+V_i(\chi_T)\in L^1([0,1]^{n-1}). $$ That implies that $\chi_S\chi_T=\chi_{S\cap T}$ has bounded variation so $S\cap T$ is a Caccioppoli set.
I assumed here that the sets are contained in the unit cube, but the argument applies to any bounded set.
Another answer is provided in a comment by Manfred Sauter (see above).