Compact subsets of infinite projective space

Let's define infinite projective space space $$\mathbb{CP}^{\infty}$$ as direct limit $$\lim \limits_{\rightarrow} \mathbb{CP}^n$$. In a class I attended it was claimed that every continuous map $$S^k \to \mathbb{CP}^{\infty}$$ is actually valued in some subspace $$\mathbb{CP}^N$$ for some sufficiently high $$N$$. How to prove this?

• A subspace of a CW-complex is compact iff it is closed and it intersects finitely many cells, see for example this question – Alessandro Codenotti Jan 31 at 19:01
• If $f_i: X_i \to X_{i+1}$ is a sequence of inclusions of compact Hausdorff spaces, then the direct limit topology on the union is such that any compact subset lies in one of the $X_i$. This is a nice exercise with the definitions. – user98602 Jan 31 at 19:06
• @MikeMiller this is precisely what I expected, but I had some trouble with verifying this property. Could you prove some additional hints? – Blazej Jan 31 at 20:04

Let $$X_i \to X_{i+1}$$ be a sequence of embeddings of $$T_1$$ spaces. Then any compact subset $$K\subset \mathrm{colim}_i X_i$$ is actually contained in an $$X_i$$.
• Thank you! I hoped for something exactly of this type, but I expected much stronger assumptions to be necessary, e.g. each $X_i$ compact Hausdorff. – Blazej Feb 1 at 21:42