Sequential compact set in $\mathbb R^2$ The problem : Show that the set of points in $\mathbb R^2$ of the form $(x,\sqrt{x})$ for
$x \in [0, 2]$ is a sequentially compact subset of $\mathbb R^2$.
My approach is to show that I can find a subsequence that converges to a point in the set.
Let $S$ = {$(x,\sqrt{x}):x \in [0,2]$}. First observe that $S$ is bounded below by $(0,0)$ and above by$(2,\sqrt{2})$.So rewrite $S$ as the interval $[(0,0),(2,\sqrt{2})]$. Let $(y_n)$ be a convergent sequence in $S$, so $y_n = (x,\sqrt{x})$. Now since every subsequence of a convergent sequence has the same limit point, it follows that the subsequence has a limit point in S.
Would this be a good solution to the problem or am I leaving anything out?
Thank you for your feedback and hints.

Let’s try to be more general.
You have a compact interval $[a,b] \subset \mathbb R$ and a continuous map $f : [a,b] \to \mathbb R$. In your case $[a,b]=[0,2]$ and $f(x) = \sqrt x$.
You have to prove that  $S= \{(x,f(x)) \ ; \ x \in [a,b]\}$ is sequentially compact.
So suppose that a sequence of points of $S$, namely $(P_n)=(x_n,f(x_n))$ converges to $(r,s) \in \mathbb R^2$. We have to prove that $(r,s) \in S$.
The coordinates of $(P_n)$ converge. So $x_n \to r$ and $f(x_n) \to s$. Continuity of $f$ implies that $f(x_n) \to f(r)$. And unicity of a limit that $s=f(r)$. Which means that $(r,s)\in S$.
We’re done.
