Problem regarding the connectedness of the Topologist's Sine Curve $$X=A\cup B=\{(x,\sin({\pi \over x}): 0\lt x\le 1\} \cup \{(0,y) : -1\le y\le 1\}\ \subset \mathbb R^2$$
is called  the  Topologist's Sine Curve - I . 
Now what  is  proved  is  that  $X$  is  connected  but  not path-connected .  
The  proof  goes  like  this :
If possible let there is a path $$\gamma:  [0,1]\rightarrow X$$ connecting the points $(0,0)$ to $(1,0)$.  Write  $$\gamma(t)=(\gamma_1(t),\gamma_2(t))$$. Then  I  can  safely  say  that  $$\gamma_1=\pi_1 \circ \gamma$$ $$\gamma_2= \pi_2 \circ \gamma$$
Since  $B$  is  a  closed  set , $\gamma^{-1} (B)$  is  also  closed in  $[0,1]$  and  bounded  also. So, by  LUB  axiom  ,  must  have  a  least  upper  bound  say  $t_0$ where  $0\lt t_0\lt 1$ .  For  being  closed  , $t_0 \in \gamma^{-1} (B)$
The  claim  is  that  $\gamma_2$  is  not  continuous  at  $t_0$ .
Choose  a  $\delta $ s.t $t_0  +\delta \le 1$ .  $$\color{fuchsia}{\text{Then } \gamma_1(t_0 +\delta)\gt0}$$
$$\color{fuchsia} {\text{Hence there exists a }n\in  \mathbb N\text{ s.t. } \gamma_1 (t_0) < { 2\over {4n+1}} < 1}$$
 By applying IVT  to  the  continuous  function  $\gamma_1$ ,  we  can  find  a  $t$  s.t  $$t_0\lt t\lt t_0 +\delta $$  satisfying $$\gamma_1 (t)={2\over  {4n+1}}$$
Then since  $$(\gamma_1(t),\gamma_2(t))=(x, sin{\pi \over x})$$ we  have  $$\gamma_2(t)=1.$$
Then $$\color{fuchsia}{|\gamma_2(t)-\gamma_2(t_0)|\ge 1}.$$
So we conclude that $\gamma_2$ is  not  continuous  at  $t_0$ .
Hence  the  $\gamma(t)$  path  does   not  exist.
Now  I  think  I  understand  all  other  parts  except  for  the  three  $\color{fuchsia}{coloured}$  steps i.e   how  those  three  steps  are  found  from  their  previous  steps  is  not  clear   to  me .
Please  help with  explanation  with  those  steps.
Thanks. 
 A: First color:
$$\color{fuchsia}{Then\ \  \gamma_1(t_0 +\delta)\gt0}$$
This follows from the fact that $t_0$ is the LUB of $\gamma^{-1}(B)$. hence, any $s>t_0$  cannot be such that $\gamma(s) \in B$.  Then, it must at $A$, where the first coordinate (namely, $\gamma_1$) will be greater than $0$.
Second color: 
$$\color{fuchsia} {\text{Hence there exists a }n\in  \mathbb N\text{ s.t. } \gamma_1 (t_0) < { 2\over {4n+1}} < 1}$$
This must be a typo. I think he meant:
$$ {\text{Hence there exists a }n\in  \mathbb N\text{ s.t. } \gamma_1 (t_0) < { 2\over {4n+1}} < \gamma_1(t_0+\delta)}$$
and this follows since $\frac{2}{4n+1} \rightarrow 0$.
Third color:
$$\color{fuchsia}{|\gamma_2(t)-\gamma_2(t_0)|\ge 1}.$$
I think something is missing here. He doesn't know the value of $\gamma_2(t_0)$. We could fix this by dividing into cases: if $\gamma_2(t_0)> 0$, choose the $n$-stuff in the second color step as to make the $\sin (t)=-1$. If $\gamma_2(t_0)<0$, choose the $n$-stuff in the second color step as to make $\sin(t)=1$.
