Is the set $\{(x,y) \in \mathbb R^2 \mid 0 < |x| \le 1 \land 0 \le y \le x^2 \} $ compact?

Question

I need to check if $$S =\{(x,y) \in \mathbb R^2 \mid 0 < |x| \le 1 \land 0 \le y \le x^2 \} $$ is compact.
I wanted to use the fact that if a set is compact, then its complement is open. Now, this point: $(0,0)$ is in $S^c$. Now, if we want to take any neighborhood this point, it will contain points from $S$.
Points in $S$ make up the area under the curve $y = x^2$ on the interval $[-1, 1]$, and so if we tried to move left or right from the origin, we will encounter points from $S$.
Is it enough to conclude that $S$ is not compact?

Answer 1

Yes it is enough to conclude that $S$ is not compact.

One way to write it is to find a sequence of points in $S$ converging to $(0,0)$.

The sequence $\left(\frac{1}{n},0 \right)$ is such a sequence.

Answer 2

Yes, here is your argument slightly more formally: any open ball centred at $(0, 0)$ contains points from $S$, since for example $(\epsilon, 0) \in S$ for any small $0< \epsilon < 1$. This means that $(0, 0)$ is in the closure of $S$. But since $(0, 0)$ is not in $S$, that means that $S$ is not closed, and hence not compact.