Let $||.||: \mathbb R^{n} \to \mathbb R, n \in \mathbb N.$
We proved that $||.||$ is continuous in the metric space $(\mathbb R^{n}, ||.||_{2})$.
Now $S_{||.||_{2}}^{n-1}:=\{x\in\mathbb R^{n}:||x||_{2}=1\}$.
And then our professor asserts that since $\{1\}$ is closed space then $S_{||.||_{2}}^{n-1}$ must also be closed, because of the above continuity $(*)$.
My problem with understanding the last line is that:
We have only proven that $||.||$ is continuous in $(\mathbb R^{n},||.||_{2})$, but surely in order for $(*)$ to be true we first need to prove that $||.||_{2}:\mathbb R^{n}\to\mathbb R$ is continuous on $(R^{n},||.||_{2}$ or $||.||)$. In other words the continuity of $||.||: \mathbb R^{n} \to \mathbb R$ on $(\mathbb R^{n}, ||.||_{2})$ has nothing to do with $S_{||.||_{2}}^{n-1}$?
What am I missing or misunderstanding here?
Any help would be greatly appreciated.
You are correct, and your professor was wrong (or you misinterpreted what they were asserting). To conclude that $S_{||.||_{2}}^{n-1}$ is closed (with respect to the $\|\cdot\|$ topology), you would instead need to know that $\|\cdot\|_2$ is continuous on $(\mathbb{R}^n,\|\cdot\|)$.
It is in fact true that $\|\cdot\|_2$ is continuous and therefore $S_{||.||_{2}}^{n-1}$ is closed, but this does not at all follow from what you have written and a separate argument is needed for it. This follows from the fact that all norms on $\mathbb{R}^n$ are equivalent, for instance, so the topology given by any two norms is the same. This means that since $S_{||.||_{2}}^{n-1}$ is closed with respect to $\|\cdot\|_2$, it is also closed with respect to $\|\cdot\|$.