The following is contained in Example 6.25 in p.289 of Dold's Lectures on Algebraic Topology.
Proposition. If $X$ is a nonorientable connected $n$-manifold then $H^n_c(X;\Bbb Z)=\Bbb Z_2$.
Proof) Consider the direct system $\{K\}$ of connected compact subsets of $X$ such that $X$ is not orientable along $K$. Then $H^n_c(X;\Bbb Z)=\text{colim} H^n(X,X-K;\Bbb Z)$. By universal coefficient theorem $H^n(X,X-K;\Bbb Z)=\text{Hom}(H_n(X,X-K;\Bbb Z),\Bbb Z)\oplus \text{Ext}(H_{n-1}(X,X-K;\Bbb Z), \Bbb Z)$. $H_n(X,X-K;\Bbb Z)=0$ by Corollary 3.4 and the torsion subgroup of $H_{n-1}(X,X-K;\Bbb Z)$ is $\Bbb Z_2$ by Corollary 3.5. Thus $H^n(X,X-K;\Bbb Z)=\Bbb Z_2$ and the result follows.
Question: How do we know that $H^n(X,X-K;\Bbb Z)=\Bbb Z_2$? Since we don't know that $H_{n-1}(X,X-K;\Bbb Z)$ is finitely generated, don't we know that $\text{Ext}(H_{n-1}(X,X-K;\Bbb Z), \Bbb Z)=\Bbb Z_2$?