Proof that a subset of a finite normed vector space is compact if and only if it's closed and bounded I'm trying to understand the proof of the following statement:

Let $(V, \|.\|)$ be a finite dimensional normed vector space. Then a subset $A \subset V$ is compact if and only if $A$ is closed and bounded.

For the first implication "$\Rightarrow$", the teacher said "it's clear". I think it comes from the fact that a norm induces a metric, which means a normed space is a metric space, and therefore we can use a proposition that says: "If $(X,d)$ is a metric space and if $A \subset X$ is compact, then $A$ is bounded and closed." But I'm not quite sure about this, so my first question is: does this "proof" make sense (for the first implication)?
Considering the second part of the proof, that is, the second implication "$\Leftarrow$", our professor wrote (translated from french, sorry if it's full of mistakes):

Let $n \in \mathbb{N}$ the dimension of $V$ and let $T: \mathbb{R}^n \rightarrow V$ an isomorphism.
$\Rightarrow \||v\|| := \|T(v)\|$ defines a norm on $\mathbb{R}^n$ and is equivalent to the Euclidean norm |.| on $\mathbb{R}^n$.
Furthermore, $T: (\mathbb{R}^n, \||.\||) \rightarrow (V, \|.\|)$ is an isometry (it preserves the distances): $\||v\|| = \|T(v)\|$.
Let $A \subset V$ a closed and bounded subset.
$\Rightarrow T^{-1} (A)$ is bounded and closed with respect to $\||.\||$ and with respect to $|.|$
$\Rightarrow T^{-1} (A)$ is compact with respect to $\||.\||$ and with respect to $|.|$
$\Rightarrow A = T(T^{-1} (A))$ is compact as well. $\square$

I don't understand why $T^{-1}(A)$ is bounded, and I'm not sure to understand what does it mean to be "closed with respect to a norm" ("par rapport à une norme" in french). I don't really understand either how to use the fact that $T$ is an isometry to make sense of all this. I managed to prove that $\||.\||$ is a norm, I understand the last implication (it comes from the fact that $T$ is a linear map from a finite dimensional space, therefore it's continuous and the result follows), but I'm a little bit lost for the rest.
Any help would be greatly appreciated!
 A: For the first implication, yes you are using that theorem. If you have a normed space, then $d(v,w):=\Vert v-w\Vert$ is a metric, so you can apply the theorem you mentioned. 
For the second, if $A$ is bounded, then there is $C>0$ such that $\Vert v\Vert\le C$ for all $v\in A$. Now if $x\in T^{-1}(A)$, then there is $v\in A$ such that $T(x)=v$ and so $\vert\vert\vert x\vert\vert\vert=\Vert T(x)\Vert=\Vert v\Vert\le C$. This shows that $\vert\vert\vert x\vert\vert\vert\le C$ for all $x\in T^{-1}(A)$, that is, that $T^{-1}(A)$ is a bounded set (using the norm $\vert\vert\vert \cdot\vert\vert\vert$).
Now in $\mathbb{R}^n$ all norms are equivalent so there exist $c_1>0$ and $c_2>0$ such that $$\vert x\vert\le c_1 \vert\vert\vert x\vert\vert\vert\le c_2 \vert x\vert$$ for all $x\in \mathbb{R}^n$. Hence if $x\in T^{-1}(A)$, then $$\vert x\vert\le c_1 \vert\vert\vert x\vert\vert\vert\le c_1 C,$$ which shows that $T^{-1}(A)$ is a bounded set (using the Euclidean norm $\vert \cdot\vert$). 
Next since $T$ is continuous, you have that $T^{-1}(\text{open set})$ is open and $T^{-1}(\text{closed set})$ is closed. Hence, $T^{-1}(A)$ is closed. Because
$$\vert x\vert\le c_1 \vert\vert\vert x\vert\vert\vert\le c_2 \vert x\vert$$ for all $x\in \mathbb{R}^n$, if you have a ball of center $x_0$ and radius $r$ with respect to $\vert\cdot \vert$, you have that it is contained in a ball of same center and radius $c_2r/c_1$ with respect to $\vert\vert\vert \cdot \vert\vert\vert $ and it contains a ball with same  center and radius $r/c_1$ with respect to $\vert\vert\vert \cdot \vert\vert\vert $. Hence, open sets with respect to $\vert\cdot \vert$ are open with respect to $\vert\vert\vert \cdot \vert\vert\vert $ and viceversa. So the open sets do not change if you change equivalent norms. In turn closed sets do not change, and also compact sets do not change. So $T^{-1}(A)$ is closed with respect to both norms. Hence, $T^{-1}(A)$ is closed and bounded with respect to $\vert\cdot \vert$  and so compact.
Since $T$ is invertible and its inverse is continuous, you now use the fact that a continuous function sends a compact set into a compact set. 
Edit 
I am adding more details. Let $B(x,r)=\{y\in \mathbb{R}^n:\, \vert x-y\vert<r\}$ and $B_1(x,r)=\{y\in \mathbb{R}^n:\, \vert\vert\vert x-y\vert\vert\vert<r\}$. If $y\in B(x,r)$ then 
$$\vert \vert\vert x-y \vert\vert\vert\le \frac{c_2}{c_1}\vert x-y\vert<\frac{c_2}{c_1}r$$ and so $y\in B_1(x,\frac{c_2}{c_1}r)$. This shows that $B(x,r)\subset B_1(x,\frac{c_2}{c_1}r)$.
On the other hand, if $y\in B_1(x,r/{c_1})$ then 
$$\vert x-y\vert\le c_1\vert \vert\vert x-y \vert\vert\vert\le c_1\frac{r}{c_1}=r$$ and so $y\in B(x,r)$. This shows that $B_1(x,r/{c_1})\subset B(x,r)$.
Now if you have a norm, say $\vert\cdot\vert$, then with this norm you form open sets with respect to this norm, namely, union of balls of the type $B(x,r)$. On the other hand, if you use the norm $\vert\vert\vert \cdot \vert\vert\vert $, then open sets with respect to this norm are union of balls of the type $B_1(x,r)$. 
By the fact that $B_1(x,r/{c_1})\subset B(x,r)\subset B_1(x,\frac{c_2}{c_1}r)$, it follows that any open set with respect to $\vert\cdot\vert$ is also open with respect to $\vert\vert\vert \cdot \vert\vert\vert $ and viceversa. 
Hence, two equivalent norms have different open balls but same open sets. 
