Let $X$ be a set. For every $A,B,C$ subsets of $X$, you can check that
$$\frac{(A\cap B)\subseteq C}{B\subseteq (A\rightarrow C)}$$
where $A\rightarrow C$ is defined as $(A\rightarrow C):=(\mathscr{C}A)\cup C$ and $\mathscr{C}A$ is the complementary set of $A$ in $X$ and horizontal line means "if and only if", i.e. the line above is true if and only if the line below is true.
More concisely,
$$A\cap (-)\dashv A\rightarrow (-)$$
where $\dashv$ means left adjoint, $A\cap(-)$ is the function sending a subset $B$ of $X$ to $A\cap B$ and $A\rightarrow (-)$ the function sending a subset $C$ of $X$ to $A\rightarrow C$.
Being a left adjoint, $A\cap (-)$ preservs colimits, in particular it preservs unions, hence your desidered distributive law.
$\textrm{EDIT}$: let $f:X\longrightarrow Y$ be a function, $X,Y$ sets, $A$ subset of $X$, $B$ subset of $Y$. Then you have
$$\frac{f(A)\subset B}{A\subset f^{-1}(B)}$$
Proceeding as in the main part of this answer, you have that the direct image of $f$ is left-adjoint to the inverse image $f^{-1}$, thus, being a left adjoint, direct image preserves colimits, for example unions, so that you immediately prove that $f(A\cup C)=f(A)\cup f(C)$, without computing on elements ( pick an element in $A\cup C$, do the image under $f$, check this image is in the image of $A$ or in the image of $C$, conversely, pick an element which is in $f(A)$ or in $f(B)$, check that it comes from an element....), while inverse image $f^{-1}$, being a right-adjoint, preserve limits, for example intersections, so that you have an "automatic" proof that $f^{-1}(A\cap C)=f^{-1}(A)\cap f^{-1}(C)$
So two facts apparently different as your distributive law and these properties of $f$ and $f^{-1}$ appear in some sense as two different manifestations of the same phenomenon