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 $A: = \{ (x, \sin(\frac{1}{x})) \mid x \in (0, 1] \}$ and $B := \{ (0, y) \mid y \in [-1, 1] \}$.
Fix any sequence $\{a_n\} _{n \in \mathbb{N}} = \{ (x_n , y_n )\} _{n \in \mathbb{N}}$ in $A \cup B$ that converge in $\mathbb{R}^2$.
Let $a = (x,y)$ be the limit point of $\{a_n\}_n$.
Let $N_A := \{ n \in \mathbb{N} \mid a_n \in A \} $ and $N_B := \{ n \in \mathbb{N} \mid a_n \in B \}$.
Then, at least one of $N_A$ and $N_B$ is unbounded in $\mathbb{N}$.
Let us assume that the essential case  that $N_A$ is unbounded (otherwise use closeness of $B$) and $x = 0$ (otherwise use continuity of $\sin(\frac{1}{x})$) occurs.
Consider the subsequence $\{a_n\}_{n\in N_A}$.
Note that the limit of any subsequence of a convergent sequence $\{p_n\}_{n \in \mathbb{N}}$ is equal to the limit of $\{p_n\}_{n \in \mathbb{N}}$ (you know?).
Since $y_n =  \sin (\frac{1}{x_n})\in [-1 , 1]$ for all $n \in N_A$ and $[-1, 1]$ is closed, we have $y \in [-1, 1]$.
Therefore $a \in B$.
Thus $A\cup B$ is closed under limit of sequence and hence it is closed.
