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?

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.
