This question is actually formulated for cohomology but I think the proof will be dual.
Let $H_\bullet$ be a general homology theory satisfying dimension axiom and union axiom. Let $(X,A)$ be a relative CW-complex, and let $X_k$ be its $k$-skeleton. Show that the inclusion $(X_k,A) \hookrightarrow (X,A)$ induces an isomorphism $H_n(X_k,A) \to H_n(X,A)$ if $n<k$.
Here my thoughts. One usually considers the long exact sequence of the triple $(X,X_k,A)$: $$ \cdots \to H_n(X,X_k) \to H_n(X_k,A) \to H_n(X,A) \to H_{n-1}(X,X_k) \to \cdots. $$ The question is whether the outer terms are zero if $n<k$. This is simple enough if $X = X_{k+1}$. In this case, one can use excision. But what if $X$ is infinite-dimensional? If I can somehow show by induction that $H_n(X_l,X_k) = 0$ for $l>k$, then one surely could use some direct limit argument to conclude for the whole space $X$.
Update @Lee Mosher: The thing is that $X_{k+1} \setminus X_k$ is fairly simple, namely a bunch of open $(k+1)$-cells. Pick for each open $(k+1)$-cell $e_i$ some point $x_i$ in $e_i$. Then $$ H_n(X_{k+1}, X_k) \cong H_n(X_{k+1}, X_{k+1} \setminus \{x_i \mid i \in I\}) \cong H_n(\coprod_i e_i, \coprod_i (e_i \setminus \{x_i\})) \cong \bigoplus_i H_n(D^{k+1}, S^k). $$ Using a long exact sequence argument on the triple $(D^{k+1},S^k,*)$ and using relative homology, i.e. $H_\bullet(D^{k+1},*) = 0$ and $H_n(S^k,*) = \delta_{nk} H_0(*)$ (this is the Kroneker delta), shows that $H_n(D^{k+1}, S^k)$ is isomorphic to $H_0(*)$ if $n=k+1$ and zero everywhere else.