I think it's a great question because not everybody is aware of this. A consequence of Brown representability theorem (Adams's version, which is highly non-trivial and depends on homotopy groups of spheres being countable) is that any spectrum $X$ fits into an exact triangle \[\coprod_{j\in J}Z_j\longrightarrow \coprod_{i\in I}Y_i\longrightarrow X\stackrel{\delta}\longrightarrow\Sigma Z\]
where $\delta$ is a phantom map and $Y_i$ and $Z_j$ are finite CW-spectra for all $i\in I$ and $j\in J$, see Neeman's "On a theorem of Brown and Adams". Hence taking homology we obtain a short exact sequence
$$\coprod_{j\in J}h_{*}(Z_j)\hookrightarrow \coprod_{i\in I}h_{*}(Y_i)\stackrel{p}\twoheadrightarrow h_{*}(X).$$
You can (and do) take without loss of generality $\{Y_i\}_{i\in I}$ to be the family of finite subcomplexes of $X$.
We also have by definition an exact sequence (not injective on the left)
$$\coprod_{\{Y_k\subset Y_i\}}h_{*}(Y_k)\longrightarrow \coprod_{i\in I}h_{*}(Y_i)\stackrel{q}\twoheadrightarrow \operatorname{colim}_{i\in I}h_{*}(Y_i).$$ The previous surjection $p$ factors throug $q$ and the canonical map $\operatorname{colim}_{i\in I}h_{*}(Y_i)\rightarrow h_*(X)$.
The first map is induced by obvious vertical map in the diagram below, and the diagonal factorization exists by general properties of triangulated categories $$ \begin{array}{ccccc} &&\coprod_{\{Y_k\subset Y_i\}}Y_k&&\\ &\swarrow&\downarrow&&\\ \coprod_{j\in J}Z_j&\longrightarrow &\coprod_{i\in I}Y_i\longrightarrow& X \end{array} $$ It is only left to show that the diagonal map is surjective on homology. This map actually has a splitting in the stable homotopy category (see Neeman's paper), hence we're done.
All this is very specific to the stable homotopy category, Adams's theorem is seldom satisfied elsewhere, hence model category arguments would usual fail in proving this.
