The set $A=\{ x :\|x\|=1 \}\subset \mathbb{R}^n$ is compact in $\mathbb{R}^n$ under euclidean norm. The set $A=\{ x \mid \|x\|=1 \}\subset \mathbb{R}^n$  is compact in $\mathbb{R}^n$ under Euclidean norm.
If I can show this is closed and bounded then we are done. It seems bounded but I am not sure how to show it is closed.
 A: Just note that every norm is continuous (because 1-lipschitz from the second triangle inequality $|\,||x||-||y||\,|\leq||x-y||$) and then $A=||\cdot||^{-1}(\{1\})$ is a closed set as a reciprocal image of the closed set $\{1\}$ of $\mathbb{R}$ by the continuous map $||\cdot||.$
A: Edit: The following answers  a former version of the question, referring to a subset of $S^{n-1}$.
The statement in the title, appearing again in the body of the question, is wrong. A subset $$A\subset S^{n-1}\subset {\mathbb R}^n$$ is compact iff it is closed. Consider, e.g., the set
$$A:=\bigl\{x\in{\mathbb R}^n\>\bigm|\>\|x\|=1, \ x_n>0\bigr\}\ .$$
The point $(1,0, \ldots,0)$ is an accumulation point of $A$, but does not belong to $A$.
A: I assume you are not familiar with the topological concept of continuity (that's the answer everyone thinks about). Now, obviously, $A$ is bounded so it suffices to prove that it is in fact closed, that is, its complement is open.
Let $p\in\mathbb R^n\setminus A.$ We shall prove that $p$ is an interior point of $\mathbb R^n\setminus A.$  
As $p\notin A$ then $\|p\|\neq1.$ Define $\varepsilon:=|\|p\|-1|.$ Then the open ball $B_{\varepsilon}(p)$ is contained in $\mathbb R^n\setminus A$ (here I used the triangle inequality) Therefore $p$ is an interior point of the complement of $A$ which means that $\mathbb R^n\setminus A$ is open and hence $A$ is closed. Thus $A$ is compact.
