Show that $\{(x,\sin(1/x)) : x∈(0,1] \} \cup \{(0,y) : y ∈ [-1,1] \}$ is closed in $\mathbb{R^2}$ using sequences Usual metric.
The abridged definitions we were just given are:
Convergent Sequence: A sequence $(x_n)$ in a metric space $X$ converges to $x_0$ if for every $\varepsilon \gt 0$ there is $n_0\in\mathbb{N}$ such that $d(x_n, x_0) \lt \varepsilon$ for each $n \geq n_0$.
Closed set: A set $F$ in a metric space $X$ is closed if and only if every convergent sequence in $F$ converges to a point in $F$.

This is the last question of a problem set to study for class. The definitions we are using are very standard for an Intro to Topology course, I just don't understand how to apply them properly yet. Would anyone be able to lend me a hand? Or show me how to start?

Let $S$ denote the set in question and fix $z \in S$.
First suppose that $z = (x,\sin(\frac{1}{x}))$ for some $x \in (0,1]$. Pick your favourite sequence $(x_n)$ such that $x_n \to x$ and $x_n \in (0,1]$ (eg. $x_n = x + \frac{1}{n}$ for large enough $n$.) ​Since $\sin(\frac{1}{x})$ is continuous on $(0,1]$ it follows that $\sin(\frac{1}{x_n}) \to \sin(\frac{1}{x})$. Hence, $(x_n,\sin(\frac{1}{x_n}))$ is a sequence in $S$ which converges to $z$.
Now suppose that $z = (0,y)$ for some $y \in [-1,1]$. Since the function $\sin$ is continuous, an Intermediate Value Theorem argument implies that for each $n \in \mathbb N$, there exists some $a_n \in [n,n+2\pi]$ such that $\sin(a_n) = y$. Let $x_n = \frac{1}{a_n}$ and observe that $x_n \in (0,1]$. Then $x_n \to 0$ and $\sin(\frac{1}{x_n}) \to y$. It follows that $(x_n,\sin(\frac{1}{x_n}))$ is a sequence in $S$ converging to $z$.
Since every point of $S$ is the limit point of some sequence in $S$, it follows that $S$ is closed.
