Why $x^2+2y^2=22$ is compact? For the set $$S=\left \{ (x,y) \in \mathbb{R}^2 : x^2+2y^2=22 \right \}$$ this is an ellipse and it looks like this:

I need to show that it is compact which means it is bounded and closed.
What is formal way to show compactness?
 A: Bounded:
Taking $s=(x,y)\in S$, one has 
$\lVert s \rVert^2 = x^2 + y^2 \leq x^2 + 2y^2 = 22$.
Closed: The function $f\colon (x,y)\in \mathbb{R}^2\mapsto x^2 + 2y^2$ is continuous, and $\{22\}$ is a closed set in $\mathbb{R}$ (can you see why both these statements hold?). Therefore, $S = f^{-1}(\{22\})$ is closed.
A: As $f(x,y) = x^2 + 2y^2$ is a continuous function, the preimage $S = f^{-1}(\{22\})$ of the closed set $\{22\}$ is closed by continuity.
Furthermore, if $x^2 + 2y^2 = 22$, then $\|(x,y)\|^2 = x^2 + y^2 \leq x^2 + y^2 + y^2 = 22$, so the set $S$ is bounded by $\sqrt{22}$. That is $(x,y) \in S \Rightarrow \|(x,y)\| \leq \sqrt 22$.
This is a standard real-analysis problem and practicing this type of problem might score some easy points on an exam.
A: $S = f^{-1}[\{22\}]$ for a suitable continuous function $f$ from $\mathbb{R}^2$ to $\mathbb{R}$, which makes it closed.
Can $|x| > 10$ or $|y| > 10$ occur?  If not (prove it), boundedness follows....
A: Consider the disk $x^2+y^2\leq 22$. Clearly it contains that elipse, so the elipse is bounded.
To see it is closed note that $f:\mathbb R^2\rightarrow \mathbb R$ defined by $f(x,y)=x^2+2y^2$ is a continuous function, so the pre-image of the closed set $\{22\}$ is closed.
