One of the lemma in May's Concise course in algebraic topology is the following:
If $X$ is the colimit of a sequence of inclusions $X_i\to X_{i+1}$ of based spaces, then the natural map $$\operatorname{colim}_i\pi_n(X_i)\to \pi_n(X)$$ is an isomorphism for each $n$. Here, we assume all the topological spaces are compactly generated.
Proof of this lemma is just one line: This follows directly from the point-set topological fact that if $K$ is a compact space, then a map $K\to X$ has image in one of the $X$.
I think the proof is showing the surjectivitiy of the natural map. To me, injectivity of the natural map is not immediate. Can someone explain why?
If $f:S^n \rightarrow X_j$ and $g:S^n \rightarrow X_k$ represent two elements of $\text{colim}_i \ \pi_n (X_i)$ which are mapped to the same element in $\pi_n(X)$, that is they are homotopic in $X$, the homotopy between them, which is a map $S^n \times I \rightarrow X$, will lie in some $X_m$ ($m$ greater than $i$ and $k$) since $S^n \times I$ is compact.
This means that $f$ and $g$ are mapped to the same element in $\pi_n(X_m)$ and thus that they actually represent the same element in $\text{colim}_i \ \pi_n (X_i)$.