Using the Heine-Borel theorem to show compactness of an ellipsoid I want to show that $E$ is compact in $\mathbb{R}^3$ where
$$E=\left\{(x,y,z)\in \mathbb{R}^3\;\middle|\;\frac{x^2}{4}+\frac{y^2}{4}+\frac{z^2}{9} =1\right\}$$
Now boundedness, doesn't that require a metric? Well, if I use the Euclidean metric (maybe I need to use this metric to employ the Heine-Borel theorem), take $O=(0,0,0)$ and then $d(O,x)\lt 5$ for all $x\in E$. Hence $E$ is bounded.
Now I need to show that $E$ is closed.
How would I show closure? I can 'see' that all neighborhoods of a point will have another point within them. But I am not sure how to show this properly
 A: The set is
$$
E=\left\{(x,y,z)\in \mathbb{R}^3\;\middle|\;\frac{x^2}{4}+\frac{y^2}{4}+\frac{z^2}{9}= 1\right\}
$$
If $\|(x,y,z)\|>3$, then $(x,y,z)\notin E$, because from $x^2+y^2+z^2>9$ we deduce 
$$
\frac{x^2}{4}+\frac{y^2}{4}+\frac{z^2}{9}
\ge\frac{x^2}{9}+\frac{y^2}{9}+\frac{z^2}{9}>1
$$
and so $(x,y,z)\notin E$. Therefore $E$ is bounded.
Since $E$ is the inverse image of $\{1\}$ under the continuous function
$$
f\colon \mathbb{R}^3\to\mathbb{R},
\quad
f\colon(x,y,z)\mapsto\frac{x^2}{4}+\frac{y^2}{4}+\frac{z^2}{9}
$$
the set $E$ is closed.
Note that the Heine-Borel theorem requires using the Euclidean metric.
A: To show that the set is closed, let $\{ (x_n,y_n,z_n) \}_{n=1}^{\infty}$ be a sequence of points on the ellipsoid that converges in $\mathbb{R}^{3}$ to some $(x,y,z)$. Then $x_n \rightarrow x$, $y_n \rightarrow y$, and $z_n\rightarrow z$. Therefore,
$$
        \frac{x^{2}}{4}+\frac{y^{2}}{4}+\frac{z^{2}}{9} = \lim_{n}\frac{x_n^{2}}{4}+\frac{y_n^{2}}{4}+\frac{z_n^{2}}{9} = \lim_n 1 = 1.
$$
So $(x,y,z)$ is on the ellipsoid, which means the ellipsoid is closed.
You can show compactness using this because, if $\{ (x_n,y_n,z_n) \}$ is any sequence of points on the ellipsoid, then $\{ x_n \}$ is bounded by $2$, $\{ y_n \}$ is bounded by $2$, and $\{ z_n \}$ is bounded by $3$. So you can choose a subsequence of the points that converges in the $x$ coordinate, and then choose a subsequence of the subsequence that converges in the second coordinate (the subsequence of the subsequence will converge in $x$ automatically because a subsequence of a convergent sequence is convergent)... continuing, you obtain a subsequence $\{ (x_{n_k},y_{n_k},z_{n_k}) \}$ that converges in all three coordinates; then the argument of the first paragraph shows that the limit of this subsequence will be on the ellipsoid.
A: The closure comes from the fact that {0} is closed and the surface itself is the inverse image of {0} of the continuous function x^2/4+y^2/4+z^2/9-1. The boundedness comes from ability of enclosing the surface inside the sphere x^2+y^2+z^2=100
