Totally bounded and closed implies compact??

You are right.

For a subset $A$ of metric space $(X,d)$ to be compact, it is not enough that $A$ is totally bounded and closed (since $X$ is always closed). However, the correct assumptions to conclude that $A$ is compact are that it is totally bounded and complete. You can follow the link I included in my previous answer to a similar question to see that proof of this fact.

$B(0,1)$ is closed in the space $B(0,1)$ endowed with the subspace topology of $\mathbb{R}^n$. But this is true for every $S \subset \mathbb{R}^n$ if we endow $S$ with the subspace topology.

The requirement is that the set is closed in $\mathbb{R}^n$, not in itself.

Moreover, in a generic tolopogical space X, given $A \subset X$, the equivalence " $A$ is compact if and only if closed and totally bounded" is correct in the case the ambient space $X$ is complete. In this case every closed subspace of $X$ is also complete.