Theorem
All norms on a finite dimensional vector space $V$ are equivalent.
Proof. Omitted.
So to follow we will use the infty norm $\lVert\cdot\lVert_\infty$ given by the equation $$ \lVert\vec v\lVert_\infty=\max\{|v^i|:i=1,...,n\} $$ where $v^i$ is for each $i=1,..,n$ the $i$-th componet of $\vec v$ with respect any basis.
So let be $V$ a finite dimensional real vector space and so let be $W$ the subspace spanned by the $(n-1)$ vectors $\vec v_1,...,\vec v_{n-1}$, that is $W$ is the set of linear combinations of these such vectors. Now by standards theorems of finite dimensional vector spaces it must exist a vector $\vec v_n$ that is not linear combination of the above vectors and additionally any vector of $V$ is a linear combination of this vector and of the preceding ones, that is for any $\vec v\in V$ there must exist $n$ scalars $v^1,...,v^n$ such that $$ \vec v=v^i\vec v_i $$ where in the above identity I have used the Einstein convention. So we define the sets $X$ and $Y$ by letting that $$ X:=\{\vec v\in V:\lambda^n>0\}\,\,\,\text{and}\,\,\,Y:=\{\vec v\in V:\lambda^n<0\} $$ so that $V\setminus W=X\cup Y$ and thus if we prove that these sets are open then the statement follows. So for $\vec v\in V$ we observe that if $\vec u\in B(\vec v;v^n)$ then $$ |u^n-v^n|\le\max\{(u^i-v^i):i=1,...,n\}=\lVert\vec u-\vec v\lVert_\infty<v^n $$ so that $$ -|v^n|<u^n-v^n<|v^n| $$ from which we conclude that if $\vec v\in X$ then $\vec u\in X$ and if $\vec v\in Y$ then $\vec u\in Y$ so that these two sets are open.
So I ask if what I tried to prove is effectively true and if my arguments are correct. So could someone help me, please?
The proof seems correct, but it's overly complicated.
Let $H$ be a subspace of $V$ of dimension $n-1$. Pick a basis $\{v_1,v_2,\dots,v_{n-1}\}$ of $H$ and complete it to a basis $\{v_1,\dots,v_{n-1},v_n\}$ of $V$.
Define the linear map $f\colon V\to\mathbb{R}$ by $$ f(v_1)=f(v_2)=\dots=f(v_{n-1})=0,\quad f(v_n)=1 $$ Then the map $f$ is continuous (prove it) and its kernel is $H$, which is therefore closed. Now $$ A=\{v\in V:f(v)>0\},\quad B=\{v\in V:f(v)<0\} $$ are nonempty disjoint open sets and $V\setminus H=A\cup B$. They are not empty because $v_n\in A$ and $-v_n\in B$.