= Solution
Construct a <bipartite graph> with left vertices $A_1,\ldots,A_k$, right vertices $B_1,\ldots,B_k$, and an edge $A_iB_j$ exactly when $A_i\cap B_j\ne\varnothing$. For $S\subseteq[k]$, every point of $\bigcup_{i\in S}A_i$ lies in some $B_j$ adjacent to one of those $A_i$. Thus
$$
\bigcup_{i\in S}A_i
\subseteq
\bigcup_{j\in N(S)}B_j.
$$
Using finite additivity of <Lebesgue measure> and the equal-volume hypotheses,
$$
\frac{|S|}{k}
=\operatorname{vol}\left(\bigcup_{i\in S}A_i\right)
\leq
\operatorname{vol}\left(\bigcup_{j\in N(S)}B_j\right)
=\frac{|N(S)|}{k}.
$$
Hence $|N(S)|\geq|S|$ for every $S$. The <Hall marriage theorem> supplies a <perfect matching>, which has the form $A_iB_{\sigma(i)}$ for a permutation $\sigma$ of $[k]$. Every matched edge means precisely that $A_i\cap B_{\sigma(i)}\ne\varnothing$.
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