$H^n_c(X)\cong H^n(X^+,\infty)$ where $X^+$ is the one-point compactification The following exercise is taken from section 3.3 of Hatcher.

21. For a space $X$, let $X^+$ be the one-point compactification. If the added point, denoted $\infty$, has a neighborhood in $X^+$ that is a cone with $\infty$ the cone point, show that the evident map $H^n_c(X;G)\to H^n(X^+,\infty; G)$ is an isomorphism for all $n$.

This is how far I have gotten:
By definition, $$H^n_c(X)=\operatorname{colim}_KH^n(X,X-K)$$
ranging over $K\subset X$ compact.
So let $U\subset X^+$ be the neighborhood which is a cone with cone point $\infty$. In particular $U^c\subset X$ is compact, and since $X-U^c=U-\{\infty\}$, we have a map $\phi:H^n(X,U-\{\infty\})\to H^n_c(X)$. By a Mayer-Vietoris argument, $$H^n(X,U-\{\infty\})\cong H^n(X^+,U)$$
and using the long exact sequence and the Five-Lemma, $$H^n(X^+,U)\cong H^n(X^+,\infty).$$
Therefore it would be enough to show that $\phi$ is an isomorphism. How can we see that? Is this the right approach?
 A: Hatcher observes that singular homology has compact carriers which is made precise in Proposition 3.33. This is an immediate consequence of the definition of singular homology.
In contrast, singular cohomology does not have compact carriers. This motivates the construction of a variant of singular cohomology denoted by singular cohomology with compact support (or carriers) by considering only cocycles with compact support. This yields a cochain complex $C_c^\ast(X;G)$ whose cohomology groups $H^n(X;G) = H^n(C_c^\ast(X;G))$ are what we desire.
An alternative description can be given via the direct limit. Let $\mathfrak{K}(X)$ resp. $\mathfrak{K}'(X)$ denote the direct system of all compact subsets $K \subset X$ resp. all subsets $K' \subset X$ having compact closure, both partially ordered by inclusion. Note that $\mathfrak{K}(X)$ is a cofinal subsystem of $\mathfrak{K}'(X)$. Then
$$H^n_c(X;G) = colim_{K \in \mathfrak{K}(X)} H^n(X,X-K;G) = colim_{K' \in \mathfrak{K}'(X)} H^n(X,X-K';G) .$$
Let $X$ be locally compact. If $\infty$ has arbitrarily small contractible neighborhoods in $X^+$ (which means that each neigborhood $U$ of $\infty$ contains a contractible neigborhood $V$ of $\infty$), we say that $X$ satisfies the ASCN-condition. Note that we do not require neigborhoods to be open. If we want open neighborhoods $U$, we get the open ASCN-condition.
We can generalize Hatcher's Exercise 21 as the follows:
If $X$ is locally compact and satisfies the ASCN-condition, then the evident map $\phi : H^n_c(X;G) \to H^n(X^+,\{\infty \};G)$ is an isomorphism.
As in Yilong Zhang's comment we see that for a contractible neighborhood $U$ the map $\phi_U : H^n(X,U - \{ \infty \};G) \to H^n(X^+,\{\infty \};G)$ is an isomorphism. But now $K'_U  = X^+ - U \in \mathfrak{K}'(X)$, and by the ASCN-condition these $K'_U$ form a cofinal subsystem of $\mathfrak{K}'(X)$, which proves the above claim.
The existence of a "cone neighborhood" of $\infty$ is nothing else than a convenient criterion assuring that the ASCN-condition holds. Note that Hatcher is not really  precise when he requires that "$\infty$ has a neighborhood in $X^+$ that is a cone with $\infty$ the cone point". A cone has the form $CZ$ with some space $Z$, and obviously Hatchers requirement means that there exists a neigborhood $V$ of $\infty$ such that $(V,\infty) \approx (CZ,\ast)$ as pointed spaces. To conclude that the ASCN-condition holds we need the requirement that $Z$ is compact, i.e. that $V$ is a compact neighborhood. 
