Topologist's Sine Curve: A rigorous proof of why $\overline{S} = S \cup V$ Let $S$ and $V$ be the following subsets of $\mathbb{R}^2$. 
$$ S = \left\{ \ x \times \sin \frac{1}{x} \ \colon  \ 0 < x \leq 1 \ \right\}, \qquad V = \left\{ \ 0 \times y \ \colon \ -1 \leq y \leq 1 \ \right\} = \{ 0 \} \times [-1, 1]. $$
Then how to show rigorously that the closure $\overline{S}$ of $S$ is given by 
$$ \overline{S} = S \cup V? $$
By rigorously, I mean using the smme kind of argument as is used in real analysis or topology, such as in (most of) the proofs in Rudin or Munkres. 
Intuitively, it is clear that as $x \to 0+0$, then $\sin \frac{1}{x}$ will fluctuate its full range of $[-1, 1]$, and for any point $0 \times y$, where $y \in [-1, 1]$, there will be points of $S$ arbitrarily close to it. 
But I'm unable to rigorously show this. 
 A: Let $v\in S\cup V$. If $v\in S$ then obviously there is a sequence in $S$ convergent to $v$, namely the constant one.
Assume that $v\in V$, i.e. $v=(0, y)$ for some $y\in[-1,1]$. Since $\sin$ is onto $[-1,1]$ and periodic then there is $z\in\mathbb{R}$, $z>1$ such that $\sin(z)=y$. Let $T>0$ be the period of $\sin$ (we all know that it's $2\pi$ but I just want to show that the choice of $\sin$ is really irrelevant here, only the fact that it is continuous, periodic and onto $[-1,1]$). Put
$$x_n=\frac{1}{z+nT}$$
It follows that $1\geq x_n>0$, $x_n\to 0$ and
$$\sin\big(\frac{1}{x_n}\big)=\sin(z+nT)=\sin(z)=y$$
In particular the sequence $\bigg(x_n,\sin\big(\frac{1}{x_n}\big)\bigg)$ is fully contained in $S$ and convergent to $v$.
What we've shown is that $S\cup V\subseteq\overline{S}$. The other inclusion follows simply from the fact that $S\subseteq S\cup V$ and $S\cup V$ is closed.
A: Let $y\in [-1;1]$. Pick $y_0$ such that $\sin \left(y_0 \right)=y$. For any $k\in \mathbb{N}$, $\sin \left( y_0 + 2k \pi \right) = y$.
Now let $\epsilon >0$. Consider $k$ large enough so that $0<\frac{1}{y_0 + 2k\pi} < \epsilon$. Put $x_n= \frac{1}{y_0+2n\pi}$ for $n\geq k$. Then the distance between $0\times y$ and $x_n \times \sin (\frac{1}{x_n})$ is $\sqrt{x_n^2} = x_n$ (since it's positive), and $x_n \to 0$ as $n\to \infty$, so $0\times y$ is in the closure. We've proved one inclusion
For the reverse inclusion, assume $x\times y \in \overline{S}$.  Then if $x=0$, clearly $y\in [-1; 1]$ (is this clear to you?), and so $x\times y \in V$. 
Otherwise, we wish to prove that $x\times y \in S$. But if $x>0$, then for some $\epsilon> 0$, $x\in ]\epsilon, 1]$. Therefore, $x\times y$ is in the closure of $\{z\times \sin(\frac{1}{z}), z\in ]\epsilon; 1]\}$, so it's in the closure of $\{ z\times \sin (\frac{1}{z}), z\in [\epsilon, 1]\}$. 
But thus set is the graph of $z\mapsto \sin (\frac{1}{z})$ on $[\epsilon, 1]$, which is a continuous function defined on a compact set (with values in a Hausdorff space); and it's therefore closed. So $x\times y = x\times \sin (\frac{1}{x})$, and $x\times y \in S$.
Therefore we do get $\overline{S} = S\cup V$
A: The set of limits points of topological sine curve is {$0$} and [$-1,1$]. That is the $y$ - axis between $- 1$ and $1$, all are limit points. That is the limit points are {$0$}$\times[-1, 1]$.
And we know that closure of a set is the union of that given set and set of limit points.
Hence, your result is proved. 
