Proof that the set $S = \{(x, y)\in \mathbb{R}^2 \mid y=x^2\}$ is closed. Our professor defined that a closed set is a set whose complement is open.
Based on this definition, how do I prove that the set $S = \{(x, y)\in \mathbb{R}^2 \mid y=x^2\}$ is closed?
It makes intuitive sense to me, but I'm unable to pen down a proof.
Thanks in advance!

I give a solution using sequence characterization:
$A\subseteq \mathbb{R}^n$ is closed if and only if for each sequence $(x_n)\subseteq A$ such that $x_n \to x$ then $x \in A$.
Let $z_n=(x_n,y_n)$ sequence in $S$ such that $z_n \to (x,y)$. So $y_n=x_n^2, \forall n\in \mathbb{N}$.
How $z_n \to (x,y)$ we have $x_n \to x$ and $y_n \to y$, then
$$ y = \lim y_n = \lim x_n^2 = x^2 $$
so $(x,y)\in A$.
