Is the intersection of two Caccioppoli (i.e. finite perimeter) sets Caccioppoli?

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).

The intersection of two Caccioppoli sets is again a Caccioppoli set: as Piotr Hajlasz points out, a Caccioppoli set is simply a set of finite perimeter therefore the (standard) proof I report below is based on the following, slightly more general, definition of the (relative) perimeter of a set.

Definition. Let $S$ a Lebesgue measurable set in $\mathbb{R}^n$. For any open subset $\Omega\subseteq\mathbb{R}^n$ the perimeter of $S$ in $\Omega$, denoted as $P(S,\Omega)$, is the variation of $\chi_S$ in $\Omega$ i.e. \begin{split} P(S,\Omega)&=\sup\left\{\int_S \mathrm{div}\varphi\,\mathrm{d}x\,:\,\varphi\in [C_c^1(\Omega)]^n, \|\varphi\|_\infty\leq1\right\}\\ & =\| \nabla (\chi_{S\cap\Omega})\|_{TV}=TV(S,\Omega) \end{split} The perimeter so defined is a lower-semicontinuous set function which is additive respect to the argument $\Omega$.

Theorem. Let $S$ and $T$ two Lebesgue measurable sets in $\mathbb{R}^n$: then, for any open set $\Omega\subseteq\mathbb{R}^n$ $$ P(S\cup T,\Omega)+P(S\cap T,\Omega)\leq P(S,\Omega) + P(T,\Omega) $$ Proof. We can choose two smooth sequences of functions $\{s_k\}, \{t_k\}$ in $C^\infty(\Omega)$ converging respectively to $\chi_S$ and $\chi_T$ in $L^1_{loc}(\Omega)$, for example by using partitions of unity, and such that $$ 0\leq s_k\leq 1\quad 0\leq t_k \leq 1, $$ and $$ \lim_{k\to\infty}\int_\Omega |\nabla s_k|\mathrm{d}x=P(S,\Omega)\quad \lim_{k\to\infty}\int_\Omega |\nabla t_k|\mathrm{d}x=P(T,\Omega) $$ Now since $s_kt_k\to\chi_{S\cap T}$ and $s_k+t_k-s_kt_k\to\chi_{S\cup T}$, the theorem follows by passing to the limit for $k\to\infty$ in the following elementary inequality $$ \int_\Omega |\nabla (s_kt_k)|\mathrm{d}x+\int_\Omega |\nabla (s_k+t_k-s_kt_k)|\mathrm{d}x \leq \int_\Omega |\nabla s_k|\mathrm{d}x+ \int_\Omega |\nabla t_k|\mathrm{d}x\quad\blacksquare $$ An immediate consequence of this theorem is the following corollary, which includes the sought for answer to the posed question.

Corollary. If $S$ and $T$ are Caccioppoli sets, so are $S\cup T$ and $S \cap T$.

Notes

• The proof is taken almost verbatim from the wonderful book of Ambrosio, Fusco and Pallara ([1], §3.3 p. 144): De Giorgi, Colombini and Piccinini ([2], §1.3 pp.18-19) prove the same result in a different way and assuming from the start that $S$ and $T$ are Caccioppoli sets.
• A direct proof of the result $P(S),P(T)<\infty\Rightarrow P(S\cap T)<\infty$ is presented by Gurtin, Williams and Ziemer ([3], p. 6) and is based on lattice and measure theoretical considerations: in fact, it is worth noting that this property of Caccioppoli sets make them very important in the formal definition of the concept of body in the modern rational continuum mechanics.
• It is perhaps also worth to point out that the property of BV functions used by Piotr Hajlasz is in reality their original definition as conceived by Lamberto Cesari back in 1936.

[1] Ambrosio, Luigi; Fusco, Nicola; Pallara, Diego (2000). Functions of bounded variation and free discontinuity problems. Oxford Mathematical Monographs. New York and Oxford: The Clarendon Press/Oxford University Press, New York, pp. xviii+434, ISBN 0-19-850245-1, MR1857292, Zbl 0957.49001.

[2] De Giorgi, Ennio; Colombini, Ferruccio; Piccinini, Livio (1972), Frontiere orientate di misura minima e questioni collegate [Oriented boundaries of minimal measure and related questions], Quaderni (in Italian), Pisa: Edizioni della Normale, pp. 180, MR 0493669, Zbl 0296.49031.

[3] Gurtin, M. E.; Williams, W. O. & Ziemer, W. P., "Geometric measure theory and the axioms of continuum thermodynamics", Archive for Rational Mechanics and Analysis, 1986, 92, 1-22, MR 816619, Zbl 0599.73002.

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.