Hawaiian Earring 
Let $X=[0,1]$ and $A=\{0\}\cup\{\frac{1}{n}|n\in\mathbb Z\}$. Note that $(X,A)$ is not a good pair. Show that $H_1(X,A)$ is not isomorphic to $H_1(X/A)$.

I have a sequence of homology groups:
$\cdots\rightarrow H_1(A)\rightarrow H_1(X)\rightarrow H_1(X,A)\rightarrow H_0(A)\rightarrow H_0(X)\rightarrow H_0(X,A)\rightarrow0$
$X$ is path-connected, so $H_0(X)=\mathbb Z$.
$A$ is nonempty and I have managed to prove that $H_0(X,A)=0$.
EDIT: As mentioned: $H_1(X)=0$ as $X$ is contractible.
I also have the following in my notes that I can't seem to justify:

*

*$H_1(X,A)$ is countable, but I don't know how to reach that conclusion.

*The map $H_1(X,A)\rightarrow H_0(A)$ is an inclusion map.

*$H_0(A)$ is $\mathbb Z<\text{uncountably many things}>$
And how do any of these help me with what I want to show?

$H_0(A)  = \mathbb{Z}\langle \text{path components of $A$}\rangle$, and since $A$ itself is countable, so must be the set of its path components (in fact every point is a path component). So in fact 3 should have countable and not uncountable. Now a countable direct sum of a countable group is countable. 
$H_1(X) = 0$, and since the sequence is exact, the next map must be an inclusion.
Hence $H_1(X,A)$ is countable.
To show that $H_1(X/A)$ has at least uncountable many $\mathbb{Z}$ independent elements, use the same argument as in example 1.25 of Hatcher. So $H_1(X/A)$ is at least $\mathbb{Z}\langle \text{uncountably many things}\rangle$. This gives you a contradiction with $(X,A)$ being a good pair.
