Showing the boundary of the unit sphere is closed I'm trying to show that this set is closed: $B=\{x\in \Bbb{R}^n \mid |x|=1\}$
Here is my proof : Let $y$ be a limit point then $B(y,r) \cap B \neq \emptyset $ then let $z\in B(y,r) \cap B \neq \emptyset$ hence $|z|=1$ . Also since  $z\in B(y,r)$ $|y-z|<r\ \forall r>0$ then $r>|y-z|>||y|-|z|| \implies ||y|-1|<r \ \forall r>0$ then $|y|=1$ hence $y\in B $. Since this is true for any $y$ B contains all its limit points and hence closed. Is this right?
 A: 
Is this right?

Yes.
Alternatively, one could note that $B=\{x\in\mathbb R^n\mid|x|=1\}=F^{-1}(\{1\})$ where $F:\mathbb R^n\to\mathbb R$ is defined by $F(x)=|x|$. Since $F$ is $1$-Lipschitz, $F$ is continuous. Since $\{1\}$ is a closed set and $F$ is continuous, $B=F^{-1}(\{1\})$ is a closed set.
A: Well it's not true to say that $|y-z|<r$ for all $r>0$ because $z$ depended upon $r$ in the previous line. However, you have the right idea that you can show that any limit point of $B$ is the limit of a sequence $x_i$ with $|x_i|=1$, and hence $|y|=1$ also.
Alternatively, if you believe that $\mathcal{O}=\{x\in \Bbb R^n \mid |x|<1\}$ is open and that $F=\{x\in \Bbb R^n \mid |x|\leq 1\}$ is closed, you can view $B$ as $F\cap\mathcal{O}^c$, which is an intersection of closed sets.
A: Nice approach. There are only a few nitpicks I'd make:

(1) After you choose your arbitrary limit point of $B$, choose an arbitrary $r>0$ (you didn't specify that), then point out that $B(y,r)\cap B\neq\emptyset$, so that we can choose our arbitrary $z\in B(y,r)\cap B$.
(2) In general, $|y-z|\geq\bigl||y|-|z|\bigr|$--we can't assume it's strict--but that doesn't really have any effect on the proof.
(3) Since $\bigl||y|-1\bigr|<r$ and $r>0$ was arbitrary, then we know $|y|=1$. Here's where the arbitrary choice of $r>0$ is important. We can't say that $|y-z|<r$ for all $r>0$ (since our choice of $z$ depended on $r$), but we can certainly say it for our arbitrary $r$, and that's enough.

