The singular homology of a disjoint union is the direct sum

The standard simplex ${\mathrm{\Delta }}^n$ is path-connected: if $u=(u_0,\dots ,u_n)$ and $v=(v_0,\dots ,v_n)$ are points of ${\mathrm{\Delta }}^n$, the straight-line map $t\mapsto (1-t)u+tv$ stays in ${\mathrm{\Delta }}^n$. Therefore any singular simplex $\sigma :{\mathrm{\Delta }}^n\to X$ has connected image. Since the sets $X_{\alpha }$ are pairwise disjoint and clopen by [L2], the image of $\sigma$ lies in exactly one summand $X_{\alpha }$.

Step 1.1 identifies $C_n(X;G)$ with the direct sum of the chain groups $C_n(X_{\alpha };G)$, degree by degree, and the singular boundary preserves the chosen summand because each face of a simplex in $X_{\alpha }$ still lands in $X_{\alpha }$. Thus $C_{\bullet }(X;G)\cong {⨁}_{\alpha \in A}{C}_{\bullet }(X_{\alpha };G)$ as chain complexes.

Cycles and boundaries of a direct-sum chain complex are taken componentwise, so homology also splits componentwise. Applying [L1] to the chain-complex isomorphism of step 2.1 yields $H_n^{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}(X;G)\cong {⨁}_{\alpha \in A}{H}_n^{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}(X_{\alpha };G)\mathrm{.}$