Closed Topologist's Sine Curve is Connected

Theorem
Let $G$ be the graph of the function $\displaystyle y = \sin \left({\frac 1 x}\right)$ for $x > 0$.

Let $J$ be the line segment joining the points $\left({0, -1}\right)$ and $\left({0, 1}\right)$ in $\R^2$.

Then $G \cup J$ is connected.

Proof
Since the open interval $\left({0 \,.\,.\, \infty}\right)$ is connected, then so is $G$ by Continuous Image of Connected Space is Connected.

It is enough, from Subset of Closure of Connected Subspace, to show that $J \subseteq \operatorname{cl}\left({G}\right)$.

Let $p \in J$, say, $\left({0, y}\right)$ where $-1 \le y \le 1$.

We need to show that $\forall \epsilon > 0: N_\epsilon \left({p}\right) \cap G \ne \varnothing$, where $N_\epsilon \left({p}\right)$ is the $\epsilon$-neighborhood of $p$.

Let us choose $\displaystyle n \in \N: \frac 1 {2 n \pi} < \epsilon$.

From Sine of Multiple of Pi Plus Half:
 * $\displaystyle \sin \left({\frac {\left({4n + 1}\right) \pi} 2}\right) = 1$

and:
 * $\displaystyle \sin \left({\frac {\left({4n + 3}\right) \pi} 2}\right) = -1$

So by the Intermediate Value Theorem, $\displaystyle \sin \left({\frac 1 x}\right)$ takes every value between $-1$ and $1$ in the closed interval $\displaystyle \left[{\frac 2 {\left({4n + 3}\right) \pi} \,.\,.\, \frac 2 {\left({4n + 1}\right) \pi}}\right]$.

In particular, $\displaystyle \sin \left({\frac 1 {x_0}}\right) = y$ for some $x_0$ in this interval.

The distance between the points $\left({0, y}\right)$ and $\displaystyle \left({x_0, \sin \left({\frac 1 {x_0}}\right)}\right) = \left({x_0, y}\right)$ is $x_0 < \epsilon$.

So $\displaystyle \left({x_0, \sin \left({\frac 1 {x_0}}\right)}\right) \in N_\epsilon \left({p}\right) \cap G$, as required.