Prove that $\overline S$ is not path connected, where $S=\{x\times \sin(\frac1x):x\in(0,1]\}$ 
The goal is to show that $\overline S$ is not path connected, where $S=\{x\times \sin(\frac{1}{x}):x\in(0,1]\}$.

Proof.
Suppose there is path $f:[a,c]\to\overline S$ beginning at the origin and ending at a point of $S$. 
The set of those $t$ for which $f(t)\in0\times[-1,1]$ is closed, so it has a largest element $b$. 
Then $f:[b,c]\to\overline S$ is a path that maps $b$ into the vertical interval $0\times [-1,1]$ and maps the other points of $[b,c]$ to points of $S$.
Replace $[b,c]$ by $[0,1]$ for convenience; let $f(t)=(x(t),y(t)).$ 
Then $x(0)=0,$ while $x(t)>0$ and $y(t)=\sin(1/x(t))$ for $t>0$. 
We show there is a sequence of points $t_n\to0$ such that $y(t_n)=(-1)^n.$ Then the sequence $y(t_n)$ diverges, contradicting the continuity of $f.$
To find $t_n$ we proceed as follows: Given $n,$ choose $u$ with $0<u<x(1/n)$ such that $\sin(1/u)=(-1)^n$. Then use the intermediate value theorem to find $t_n$ with $0<t_n<1/n$ such that $x(t_n)=u.$




*

*I don't understand how this function $f(t)=(x(t),y(t))$ is defined. $f(0)=(x(0),y(0))=(0,?)$

*About the sequence of points $t_n$, I don't understand the process of construction.
Given $n\in\mathbb N,$ let $u\in(0,x/n)\dots$ Is  $x$  is a function right?
Using IVT, why the interval for $t_n$ is $(0,1/n)$?
 A: 
$f(0) = (x(0),y(0)) = (0,?)$

It doesn't matter. Take the value of $y(0)$ to be anything you like. (From the paragraph above, $f$ simply "maps $b$ into the vertical interval $0\times [-1,1]$".)

$t_n$

The point is essentially that we want to construct a sequence of points $t_n$ converging to $0$ (so the author has chosen $0 < t_n < 1/n$ for each $n$, which guarantees that $t_n\to 0$) but such that $y(t_n)$ doesn't converge at all. The $x$ here is indeed a function - it's the same function $x(t)$, just now with $t = 1/n$.
A: This answer comes in three short sections that provide the details allowing one to unravel the 'mysteries' behind Munkres proof.

Starting alongside of Munkres, we have a continuous function
$\tag 1 f:[a,c] \to \mathbb R \times \mathbb R \text{ such that } f(a) = (0,0) \text{ and } f(c) = s_0 \text{ with } s_0 \in S$
Munkres argues that there exists a unique number $b$ such that $a \lt b \lt c$ and $f(b) \in \{0\} \times [-1, 1]$ but $f$ applied to the interval $(b,c]$ lies inside of $S$. Let the $y\text{-coordinate}$ of $f(b)$ be denoted by $b_2$. 
So we can restrict $f$ to $[b, c]$ and $f(b) = (0, b_2)$ and $f(c) = s_0$. A reparameterization can be found, so finally we have (using $f$ again):
$\tag 2 f:[0,1] \to \mathbb R \times \mathbb R \text{ such that } f(0) = (0,b_2) \text{ & } f(1) = s_0 \text{ with } s_0 \in S \text{ & } f(0,1] \subset S$
Now Munkres notes that the function $f$ in (2) can be written as $f(t) = (x(t), y(t))$. 
Question: Is there second reparameterization where $x(t) = t$? If yes, does this mean that we have a real valued continuous function 
$\quad g: [0, +\infty) \to \mathbb R \text{ such that } g(x) = sin(\frac{1}{x})
 \text{ for } x \gt 0$
The Munkres analysis can be compared with facts known about such functions $g$, defined by specifying the value $g$ takes at $x = 0$. See
$\quad$  Show $f(x)=\sin\frac1x$ is discontinuous on $\Bbb R$ using open balls

Any function $f: [0,1] \to \mathbb R \times \mathbb R$ can be written in the form $f(t) = (x(t), y(t))$. If $f$ is continuous then both $\pi_x \circ f$ and $\pi_y \circ f$ are continuous, where $\pi_x$ and $\pi_y$ are the coordinate projection maps.

The following is offered as a 'baby step' in understanding Munkres proof and might provide an 'intuitive feel' for the problem:
Now $S=\{(x\times \sin(1/x)):x\in(0,1]\}$ and for any $v \gt 0$ 
$\tag p S \; \;\; \bigcap \; \; \; (0, v] \times \{+1\} \;\;\; \ne \emptyset$
and 
$\tag n S \; \; \; \bigcap \; \; \; (0, v] \times \{-1\} \;\;\; \ne \emptyset$
To see this, note that the function $y = sin(x)$ is periodic and 'keeps hitting' both $+1$ and $-1$.
The OP in encouraged to draw some pictures here that includes the shape of $f(t)$ described by (2) in Section 1.

If $sin(1/u)=(-1)^n \text{ then } 1/u = arcsin((-1)^n)$.
So
$u = (-1)^n (\frac{\pi}{2} + 2 \pi k)^{-1}$
A: Here is a  proof in a more topological style: For brevity let $T=\bar S \setminus S.$ 
Suppose $f:[0,1]\to \bar S$ is continuous  with $f(0)\in T.$  For any $x\in [0,1]$ such that $f(x)\in T$ let $f(x)=(0,v)$ and let $$V(f(x))=\bar S\cap [(-1/2,1/2)\times (-1/2+v,1/2+v)].$$ Then $V(f(x))$ is a nbhd of $f(x)$ in the space $\bar S.$ By continuity of $f$ there exists $\delta >0$ such that $f$ maps $[0,1]\cap (-\delta+x,\delta +x)$ into $V(f(x)).$ For brevity let $$[0,1]\cap (-\delta+x,\delta+x)=U(x).$$ An inspection of the graph of $S$ shows that $V(f(x))\cap S =\cup W$ where $W$ is a countably infinite  family of pair-wise-disjoint  sets of the form $\{(d,\sin 1/d):d'<d<d''\},$ with $0<d'$,  such that $\bar w_1\cap \bar w_2=\emptyset$ for any distinct $w_1,w_2 \in W.$ (Closure bar denoting closure in $\Bbb R^2$.)
Each member of $W$ is  open-and-closed in the sub-space $V(f(x)).$  So, since  the continuous image of the connected space $U(x)$  is connected, there is at most one $w_0\in W$ such that $\emptyset \ne w_0\cap f(U(x)).$  
But  if $w_0\in W$ and $\emptyset \ne w_0\cap f(U(x))$  then the two sets $w_0\cap f(U(x))$  and $T\cap f(U(x))$ are  disjoint and not empty....  ( Note: $T\cap f(U(x))$ is not empty as it contains $f(x)\;$)....and are both open in the connected sub-space $f(U(x)),$ and their union is  $f(U(x)).$ This  contradicts the connectness of $f(U(x)).$ Therefore no such $w_0$  exists. 
So $f(U(x))\cap w$ for all $w\in C.$ So $f(U(x))\subset T.$ 
We now have a kind of Heine-Borel argument: $f(0)\in T$ and  if $f(x)\in T$ then $x$ has a  nbhd $U(x)$ in the space $[0,1]$ such that $f((U(x))\subset T,$ from which we readily prove that $f([0,1])\subset T=\bar S \setminus S.$ 
So there cannot be a  path from any member of $T$ to any member of $S.$ 
A: Let $S=\{(x, sin(1/x)) \;:\; x\in(0,1]\}$ and $T = \{0\} \times [-1, 1]$. Then
$\tag 1 \overline S \text{ is equal to the disjoint union of } S \text{ with } T$
Lemma: Let $a \lt u \lt b$ and $f: (a, b) \to \overline S$ be a continuous function such that $f(u) \in S$.Then the range of $f$ is included in $S$.
Proof (Sketch)
To arrive at contradiction assume that $f$ hits points in $T$. Now the set $S$ is open in $\overline S$ and its inverse image $U$ is therefore open and contains $u$. Moreover, the complement of $U$ is a nonempty closed subset of $(a,b)$. By employing logical symmetry, we assume now that we have a number $c \gt u$ such that $f(c) \in T$ and the image of the open interval $(u, c)$ under $f$ is contained in $S$. Since the left limit $\lim_{x \to c-}f(x)$ exists one can show that $f$ can't be going 'crazy' inside of $S$ as $x$ approaches $c$ from the left. More precisely, it can be shown that there exist an $n \ge 1$ such $f$ maps $(u,c)$ into the open set
$\tag 2 S_n =\{(x, sin(1/x)) \;:\; x\in(\frac{1}{n},1]\}$
of $\overline S$, since otherwise we can argue that $f(c) = (0,+1)$ and $f(c) = (0,-1)$. But then of course $f(c)$ must be in $S$, a contradiction. $\quad \blacksquare$
A: Let $H=\{(x, sin(1/x)) \;:\; x\in(0,1)\}$ and $T = \{0\} \times [-1, 1]$. Let
$\tag 1 S = H \cup T$
It is easy to show that $S$ is path-connected if and only if $S \cup \{\left(1,sin(1)\right)\}$ is path-connected. This is only mentioned because we modified (ever so slightly) the OP's question to fit the machinery found below.
The space $S$ is not path-connected.
Assume, to get a contradiction that we have a path $\gamma$ in $S$ connecting the point $(0,0)$ to $\left(\frac{1}{2}, sin(\frac{1}{2})\right)$.  The inverse image of $T$ under $\gamma$ is a closed subset of $[0,1]$ and has a maximum value not equal to $1$. By modifying this path, we can define another path 
$\tag 2 \omega: [0,1] \to S$ 
satisfying $\omega(0) \in T$, $\,\omega(1) = \left(\frac{1}{2}, sin(\frac{1}{2})\right)$ and $\omega\left( (0,1] \right) \cap T = \emptyset$.
But this path can also be regarded as connecting the two points inside the space 
$\tag 3 G = H \cup \{ (0, \pi_y(\omega(0)) \}$ with $\pi_y$ the projection onto the $y\text{-axis}$
The concluding remark found here means that this is impossible.
