It is not true that every map from $\mathbb CP^{\infty}$ to a finite complex is null-homotopic. For example it is a result due to Brayton-Gray that there are non-null maps $\mathbb CP^{\infty} \to S^3$, this story is told in P. May's "More Concise Algebraic Topology" (see cor. 2.4.3 on page 40). However, the examples constructed there are phantom maps, i.e. they become trivial when restricted to any finite subcomplex $\mathbb CP^k \subset \mathbb CP^{\infty}$. (A way to think about this is that it is just not possible to find a compatible choice of null-homotopies.) So my question is:
Does there exist a finite CW-complex $X$ and a non-phantom map $\mathbb CP^{\infty} \to X$?
Here are some thoughts:
(1) One can show that any such map has to be trivial on cohomology (with $\mathbb Z$ coefficients, using the cup product structure of $\mathbb CP^{\infty}$) and consequently also on homology (here we have to use that $X$ is finite). Lifting the map to the universal cover $\tilde{X}$ of $X$ and using Hurewicz, one can also see that the map is zero on all homotopy groups ($\mathbb CP^{\infty}$ only has one). However, there are plenty of non-trivial maps inducing zero on all homotopy and homology groups, the easiest example is maybe given by the composition $T^3 \to S^3 \to S^2$ where the first map collapes the complement of a ball and the second one is the Hopf map.
(2) Using the Sullivan conjecture/Miller's theorem one can probably show that the $p$-completion of any such map is null.
(3) Going in the opposite direction, maybe one can start with a non-trivial map $\mathbb CP^{k} \to X$ which induces zero on all homotopy and homology groups and then somehow prove that it extends to $\mathbb CP^{\infty}$. However I do not see how one could show in practice that extending the map is possible...
