An isomorphism between two normed vector spaces with the same finite dimension is an homeomorphism I'm trying a simple proof of this fact:
An isomorphism between two normed vector spaces with the same finite dimension is an homeomorphism.
I've tried in this way (everything seems to be ok, I ask for confirmation please):
$f:\underbrace{X}_{\dim X=m}\to \underbrace{Y}_{\dim Y=m}$ defined as $f(x^1e_1+...+x^me_m)=x^1b_1+...+x^mb_m$ where $(e_1,...,e_m)$ e $(b_1,...,b_m)$ are two basis for $X$ and $Y$ respectively.
The map $f$ is obviously linear and invertible. It only remains to show that $f$ and $f^{-1}$ are both continuous. But $f(x)=\underbrace{f_1(x)}_{X\ni x\mapsto x^1b_1\in Y}+...+\underbrace{f_m(x)}_{X\ni x\mapsto x^mb_m\in Y}$ where $f_j(x)$ is continuous because it is a bounded transformation:
$$|f_j(x)|_Y=|x^jb_j|_Y=|x^j|\cdot |b_j|_Y$$
Being $f$ the sum of $m$ continuous functions, it is continuous too.
Finally, knowing that $f^{-1}(y)$ has expression: $f^{-1}(y^ib_1+...+y^mb_m)=y^1e_1+...+y^me_m$, one can repeat the same reasoning just done to conclude that $f^{-1}$ is also continuous.
Is it all ok?
If yes, I can't understand why in these notes: https://courses.maths.ox.ac.uk/node/view_material/42040 (Theorem 4.2, pag. 27) the same problem (in the particular case in which one of the two normed spaces is $\mathbb{R}^n$) was treated in this much complicated way (there, $Q$ is my $f$ and $P$ is my $f^{-1}$):

where in particular I can't understand why he says what is highlighted in blue.
 A: Some comments on what you wrote.
First: $f_j(x)$ is continuous because it is a bounded transformation:
$$|f_j(x)|_Y=|x^jb_j|_Y=|x^j|\cdot |b_j|_Y$$
I don't see how this is a proof that $f_j$ is bounded. To prove that it is bounded, you have to prove that it exists $\alpha > 0$ such that for $x \in X$ with $\Vert x \Vert_X \le 1$, you have $\vert f_j(x) \vert \le 1$. You have to introduce the norm $\Vert \cdot \Vert_X$ in your proof.
Same regarding $f^{-1}$. The norms $\Vert \cdot \Vert_x$ of $X$ and the one $\Vert \cdot \Vert_Y$ of $Y$ have to appear in the proof.
Also you have to convince yourself that in finite dimensional vector spaces, all linear map are continuous. But this is not always the case for infinite dimensional spaces. So somewhere in your proof the fact that the spaces have finite dimensions must appear.
Regarding what is written in blue, this comes from the following arguments:


*

*The inequality $\Vert Px \Vert \le \delta^{-1} \Vert x \Vert$ is clear for $x = 0$.

*If $x \neq 0$, then as $Q$ is surjective, it exists $y \in Y \setminus \{0\}$ such that $x = Qy$. The $\Vert \cdot \Vert_\infty$ norm of $y / \Vert y \Vert_\infty$ is equal to $1$ and therefore $\Vert Qy \Vert \ge \delta \Vert y \Vert_\infty$. We also have $y = Q^{-1}x= Px$. Plugging this in previous inequality, we get the desired conclusion.

