explanation of specific proof of why topologist's sine curve is not path-connected I know this question has been asked before, but I don't want a full proof, I just want an explanation of this proof of this on the internet, (shown in the picture).
I understand all of it except for why they say the closure of $\gamma ([0,c])$ contains all of $X_1$.
 
 A: I assume that 
\begin{align*}
X_{1}  &  =\left\{  \left(  x,y\right)  \in\mathbb{R}^{2}:\,x=0,\,-1\leq
y\leq1\right\}  ,\\
X_{2}  &  =\left\{  \left(  x,y\right)  \in\mathbb{R}^{2}:\,0<x\leq\frac
{1}{\pi},\,y=\sin\frac{1}{x}\right\}.
\end{align*}
To answer your question, you first observe that
$\gamma\left(  \left[  0,t_{0}\right]  \right)  $ contains $X_{2}$.
Indeed, if there existed $0<x_{0}<\frac{1}{\pi}$ such that $\left(  x_{0}%
,\sin\frac{1}{x_{0}}\right)  \notin\gamma\left(  \left[
0,t_{0}\right]  \right)  $, then the open sets $\left(  -\infty,x_{0}\right)
\times\mathbb{R}$ and $\left(  x_{0},\infty\right)  \times\mathbb{R}$ would
disconnect $X$. 
Now if you take any point $(0,y)$ on the segment $X_1$ you can find $\theta\in [0,2\pi]$ such that $\sin\theta=y$. Since $\sin(\theta+2n\pi)=y$ for every $n$, the sequence $\{(\frac1{\theta+2n},y)\}_n$ belongs to $X_2$ and converges to $(0,y)$ as $n\to\infty$. Hence $(0,y)$ belongs to the (sequential ) closure of $X_2$.  
A: After inferring from the context that $X_1=\{0\}\times [-1,1]:$
For brevity let $S=\gamma ([0,c)).$
(i). $c>0$ : Take a nbhd $U$ of ($1/\pi ,0)$ with $U\cap X_1=\phi.$ The continuity of  $\gamma$  implies there exists $r>0$ such that $\gamma([0,r))\subset U,$ so $c\geq r>0.$
(ii). The space $S$ is a continuous image of the connected space $[0,c)$ so $S$ is a connected space.
(iii). $\gamma(c)\in$ Cl$(S)$ because $c\in$ Cl$([0,c)$ (because, by (i), $c>0$) and $\gamma$ is continuous.
(iv). If there existed $x\in (0,1/\pi)$ with  $(x,\sin 1/x)\not \in S$ then the sets $A=\{(u,v)\in S:u<x\}$ and $B=\{(u,v)\in S:u>x\}$  are open disjoint subsets of the space $S$ with $A\cup B=S$ . And $B\ne \phi$ because $(1/\pi,0)\in B.$ And $A\ne \phi$ otherwise (iii) is contradicted.  But then $S$ is not a connected space, contrary to (ii).
Therefore $S\supset \{(x,\sin 1/x):0<x\leq 1/\pi\},$ implying $\overline {\gamma([0,c])}\supset \overline S\supset X_1.$
