= Solution
Use <simplicial homology> with a fixed coefficient group or ring. The <Mayer–Vietoris sequence> is
$$
\cdots\longrightarrow H_{i+1}(X)\xrightarrow{\delta}H_i(A\cap B)
\xrightarrow{\alpha}H_i(A)\oplus H_i(B)
\xrightarrow{\beta}H_i(X)\xrightarrow{\delta}H_{i-1}(A\cap B)\longrightarrow\cdots.
$$
The inclusion maps define $\alpha([z])=([z],-[z])$ and $\beta([a],[b])=[a+b]$. To define $\delta$, split a <simplicial cycle> $c\in C_{i+1}(X)$ as $c=a+b$ with $a$ a chain in $A$ and $b$ in $B$. Since $\partial a=-\partial b$, this boundary lies in $C_i(A\cap B)$, and $\delta[c]=[\partial a]$. At degree zero the unreduced sequence ends with $H_0(A)\oplus H_0(B)\to H_0(X)\to0$.
If $[z]=\delta[c]$, its representative is a boundary in $A$ and the negative of a boundary in $B$, so $\alpha[z]=0$. Thus $\operatorname{im}\delta\subseteq\ker\alpha$.
Conversely if $\alpha[z]=0$, choose $a\in C_{i+1}(A)$ and $b\in C_{i+1}(B)$ with $\partial a=z=\partial b$. The chain $c=a-b$ is a <simplicial cycle> in $X$. Its splitting into the $A$ chain $a$ and $B$ chain $-b$ gives $\delta[c]=[z]$. Hence \b[$\operatorname{im}\delta=\ker\alpha$], the requested exactness at $H_i(A\cap B)$.
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