= Solution
The <Mayer–Vietoris sequence> for $K=M\cup N$ is the <long exact sequence in homology>
$$
\cdots\to H_n(M\cap N)\xrightarrow{(i_*,-j_*)}H_n(M)\oplus H_n(N)\xrightarrow{k_*+l_*}H_n(K)\xrightarrow{\partial_n}H_{n-1}(M\cap N)\to\cdots.
$$
We use integer coefficients. A <simplicial cycle> $c$ in $K$ can be written $c=m+n$ with chains in $M,N$. Since $\partial m=-\partial n$, this chain lies in $M\cap N$ and is a cycle. Define $\partial_n[c]=[\partial m]$; changing the splitting or the cycle representative changes this by a boundary, giving a well-defined <connecting homomorphism>.
First consider two acyclic subcomplexes. Their intersection is either empty or acyclic. If empty, positive-degree <homology (mathematics)> of their disjoint union vanishes. If nonempty, the reduced <Mayer–Vietoris sequence> implies the same conclusion, including degree one since the intersection has zero reduced $H_0$. This proves the claim for two subcomplexes.
Inductively let $U=M_1\cup\cdots\cup M_{n-1}$, $V=M_n$, and $W=U\cap V=\bigcup_{i<n}(M_i\cap M_n)$. Discard empty members as necessary. Both $U$ and $W$ satisfy the same intersection hypothesis with at most $n-1$ members, so their homology vanishes in degrees at least $n-2$. The exact segment $H_i(U)\oplus H_i(V)\to H_i(K)\to H_{i-1}(W)$ then gives $\boxed{H_i(K)=0\text{ for }i\geq n-1}$. For $n\geq3$ these degrees are positive and the induction applies directly; empty $W$ again means a disjoint union.
For sharpness, take the boundary of an $(n-1)$-simplex, covered by its $n$ facets. Every nonempty proper intersection of facets is a simplex and thus has the <homology of a point>; the intersection of all facets is empty. The union is a <sphere> $S^{n-2}$ and $H_{n-2}$ is nonzero. For $n=2$ it is two isolated vertices and $H_0\cong\mathbb Z^2$; for $n\geq3$, $H_{n-2}\cong\mathbb Z$. \b[The degree bound cannot be lowered.]
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