Topologist's sine curve is not path-connected Is there a (preferably elementary) proof that the graph of the (discontinuous) function $y$ defined on $[0,1)$ by
$$ y(x) =\begin{cases}
\sin\left(\dfrac{1}{x}\right) & \mbox{if $0\lt x \lt 1$,}\\\
0 & \mbox{if $x=0$,}\end{cases}$$
is not path connected?
 A: Assume, to get a contradiction, that the graph $G \subset \mathbb R \times \mathbb R$ of the function
$$ y(x) =\begin{cases}
\sin\left(\dfrac{1}{x}\right) & \mbox{if  $0\lt x \lt 1$}\\\
0 & \mbox{if  $x=0$}\end{cases}$$
is path-connected.
Let $\gamma$ be a path in $G$ connecting $(\frac{1}{2},\sin(\frac{1}{2}))$ to $(0,0)$. We know that the image $K$ of $\gamma$ is also compact.
The sets defined by
$\tag 1 L_n = K \cap \{(x,y) \in \mathbb R \times \mathbb R \, | \, x \le \frac{1}{n} \text{ and } y = 1\} \text{ for } n \ge 2$
define a decreasing chain of closed subsets of $K$.
Now for any $0\lt\alpha\lt\frac{1}{2}$, the path $\gamma$ must pass through $(\alpha, \sin(\frac{1}{\alpha})$. This follows since the image of $\gamma$ is connected and can't be disconnected by the two open half planes defined by $x = \alpha$. Also note that for every  $k \ge 1$, the point $\left(\sin(\frac{\pi}{2} + 2 \pi k)^{-1}), 1\right)$ belongs to $G$. So each $L_n$ must be a non-empty closed subset $K$.
Recall the following general theory for a compact topological space $X$:

Any collection of closed subsets of X with the finite intersection
property has nonempty intersection.

We know that our chain $L_2 \supset L_3 \supset L_4\supset \dots$ of closed sets in $K$ satisfies the finite intersection property, so the intersection must be nonempty.
A simple argument shows that the intersection of the $L_n$ must exclude all of $G$ except perhaps the singleton set containing $(0,0)$. But this point does not lie on the line $y = 1$, so the intersection is indeed empty. But this is absurd.

Let $\beta \in \mathbb R$. Using the argument above, we can also show that the graph of the function
$$ y(x) =\begin{cases}
\sin\left(\dfrac{1}{x}\right) & \mbox{if  $0\lt x \lt 1$}\\\
\beta & \mbox{if  $x=0$}\end{cases}$$
can't be path-connected.
Using this fact, one can show that the Topologist's sine curve as defined by Munkres is also not path-connected; see this stackexchange answer.
A: An attempt at rewriting Jonas Meyer's answer. Also I prove that closure of sine curve, which contains a whole segment on $y$-axis is not path-connected:
Let $f=(f_1,f_2):[a,b]\to S$ be any continuous map such that $f(a)=(0,0)$. Since projections $(x,y) \to x$ and $(x,y) \to y$ are continuous maps $\mathbb{R}^2 \to \mathbb{R}$, we see that both $f_1:[a,b] \to \mathbb{R}$ and $f_2$ are continuous.
By continuity at $a$ of $f_2$, there exists a $\delta > 0$ such that
$$ (*) \ \ \ \ |f_2(x) - 0| < \frac{1}{2} \ ,$$ for all $a\leq x \leq a+\delta$.
Claim: For all $a\leq x \leq  a+\delta$, $f_1(x)=0$.
Proof of the claim: Assume $f(x^*) = \tau >0$ for some $a < x^* \leq a+\delta$, $f_1(x)=0$, then since $f_1$ is continuous, by Intermediate Value Theorem $f_1([a,x^*]) \supset [0,\tau) \ $  .
Therefore there exists some $t=\frac{1}{2n\pi+\pi/2}$ in the range of $f_1([0,\delta)) \ $, i.e there exists $\tilde{x} \in [a,a+\delta]$ such that $f_1(\tilde{x})=t$ We see that $$f_2(\tilde{x}) = \sin(1/t) =1, $$
which contradicts (*) above.
This proves the claim. The claim means that $f$ cannot leave the $y$-axis for some while ($\delta$ here.)
But now applying the same argument to the continuous map $[a+\delta , b] \to S$, we see that $f$ will NEVER be able to leave the $y$-axis.$^1$ In other words, we proved that any continuous path into $S$ starting on $y$-axis remains on $y$-axis. This refutes the possibility of paths between $y$-axis points and other points outside it.
$^1$ To make this precise we define $sup$ of such delta, and then reach a contradiction if this supremum is smaller than $b$. Similar construction done by Jonas in his answer.
A: If $S=\{(0,0)\}\cup\{(x,\sin(1/x)):0<x<1\}$ and $f=(f_1,f_2):[0,1]\to S$ is a path with $f(0)=(0,0)$, then $f(t)=(0,0)$ for all $t$.  
To see this by contradiction, suppose that $f(t)$ is not always $(0,0)$.  Removing an initial part of the interval and then rescaling if necessary, assume that $0=\sup\{t:f([0,t])=\{(0,0)\}\}$.  By continuity of $f_2$, there is a $\delta>0$ such that $|f_2(t)|<1$ for all $t<\delta$.  Take $t_0$ with $0<t_0<\delta$ and $f_1(t_0)>0$.  By continuity of $f_1$ and the intermediate value theorem, $[0,f_1(t_0)]$ is in the image of $f_1$ restricted to $[0,t_0]$.  Since $f_2(t)=\sin(1/f_1(t))$ for all $t$ with $f_1(t)\neq0$ and $\sin(1/x)$ maps $]0,\varepsilon[$ onto $[-1,1]$ for all $\varepsilon>0$, it follows that $[-1,1]$ is in the image of $f_2$ restricted to $[0,t_0]$.  This contradicts $t_0<\delta$.
