Proof:

We shall only sketch the proof leaving the details as an exercise. Note that if $\sigma$ is a singular $k-$simplex, the image of $\sigma$ must be contained in one of the components $X_{\alpha}$ of $X$ and so may be regarded as a singular $k-$simplex in $X_{\alpha}$. This gives a natural decomposition of $S_k(X)$ as a direct sum of the family $S_k(X_{\alpha})$. To see that the boundary map $\partial _k$ respects the decomposition note that the boundary of a singular simplex $\sigma$ is a sum of finitely many $k-1$ singular simplexes each of which must map into the same component as $\sigma$. It is easy to deduce from this the decompositions $Z_k(X) = \bigoplus Z_k(X_{\alpha})$ and $B_k(X) = \bigoplus B_k(X_{\alpha})$.