Why do I keep proving that if $S$ is path connected so is it's closure $\bar{S}$. Topologist's sine curve is clearly a counterexample for this situation. What am I doing wrong below?
Let $A \subset X$ be path connected. Show that so is $\bar{A}.$
$x \in \bar{S}$ if and only if every open $\epsilon$ neighborhood centered at $x$, say $B_{\epsilon}(x) $ intersects $A$. 
Let $x , y \in \bar{S}$. Then assign two open neighborhoods : $B_{\epsilon}(x)$ and $B_{\delta} (y)$. Since each of these neighborhoods intersect A, then there exists two points $p,q \in A$ such that $p \in B_{\epsilon}(x)$ and $q \in B_{\delta} (y)$. Since $A$ is path connected there exists a path $P_1$ between $p$ and $q$. Also, there exists paths $P_2$ between $p$ and $x$, path $P_3$ between $q$ and $y$. 
Then $$ P = P_1 \cup P_2 \cup P_3$$
is a path connecting $x$ and $y$.
However, take the topologists sine curve:
$$S = (x \ \times \ sin(1/x) \ |  \ 0 < x \le 1)$$
which is clearly path connected with $\bar{S}$ being not.
 A: There is no reason why $P_2$ and $P_3$ should lie inside $\overline{A}$, which is the requirement to see that $\overline{A}$ is path-connected.  They just can be chosen inside the open balls. 
A: Let's start by understanding why the topologists' sine curve $S$ is actually a counterexample to the proposition. Notice that $\bar{S} = S \cup (\{0\}\times [-1, 1])$. The issue with path-connectedness is that there's no way for a path starting somewhere on $\{0\}\times[-1, 1]$ to continuously jump to the graph of $\sin(1/x)$. The graph has no good point of entry because it's oscillating so much. Precise proof is on page 6 here -- it just formalizes this intuition into a proof-by-contradiction. 
The implication for your proof is that you need to think more carefully about why you can move from your starting point $x$ to the point $p$ through $\bar{A}\cap B_\epsilon(x)$. Certainly if you leave $\bar{A}$, you can find $P_2$, but then you're just proving that $\mathbb{R}^2$ is connected, not that $\bar{A}$ is connected.
In the case of the topologists' sine curve, what does your proof say? If $x$ lies on the vertical segment on the $y$ axis, note that $\bar{S}\cap B_\epsilon(x)$ has infinitely many connected components --- the intersections of the rapid oscillations with the small $\epsilon$-disk --- and so for any $p\in S$ you will certainly not be able to find a path from $x$ to $p$ passing through $\bar{S}\cap B_\epsilon(x)$.
