I have written a proof for the following and would like you to correct me if I made any mistakes, thanks in advance:
claim: $X = \sqcup_i X_i$ then $H_q (X) \cong \oplus_i H_q (X_i)$
proof: Proof of case $X = A \sqcup B$, the general case follows by induction.
By the Mayer-Vietoris theorem the following sequence is exact:
$ \dots \xrightarrow{k_\ast} H_{n+1}(X) \xrightarrow{\partial_\ast} H_n(A \cap B) \xrightarrow{(i_\ast, j_\ast)} H_n(A) \oplus H_n(B) \xrightarrow{k_\ast} H_n(X)\xrightarrow{\partial_\ast} \dots$
Then $A \cap B = \emptyset \implies H_n(A \cap B) = 0 \implies \partial_\ast = 0$
Then $k_\ast$ is injective because $ker k_\ast = im \partial_\ast = 0$ and $k_\ast $ is surjective because $im k_\ast = ker \partial_\ast = H_n(X)$