Equivalence of norm and weak topologies in finite dimensional space. I haven't solved a lot of problems on functional analysis, so I don't have intuition whether my idea of proof is correct or not.
The weakness of weak-topology is obvious from the definition. To show another direction of the proof take an open ball $B(a,\epsilon )$ in norm topology. Since all norms are equivalent in finite dimensional spaces this could be a usual ball with standart metric. Let $y\in B(a, \epsilon)$ and $f:X\to \mathbb{R}$ s.t. $f(e)=1$ for any basis element of vector space. Then I can find $(u,v) \subset \mathbb{R}$ s.t. $y$ lies in weakly open set $f^{-1}(u,v)\subset B(a, \epsilon )$.
I know that it is not a rigorous proof(if it is indeed a proof). I'm asking about correctness of idea itself.
 A: First a few comments about your statement:
1) I assume you are thinking in a certain $f$ which is a continuous linear functional (otherwise the argument does not say anything).
2) What do you mean by "a basis element of the the vector space". If you mean any element $e$ such that $\{e\}$ can be extended to a basis of the space, then $e$ can be any non $0$ element (which is really not saying much).
3) Note that you are not really using the finite dimensional structure, what hints that there must be something wrong in the argument since the result does not hold for infinite dimensional spaces.
Indeed, what fails in your argument is that it is not true in general that you can find $(u,v)$ such that $f^{-1}(u,v)$ is included in $B(a, \epsilon)$. This is the case in any infinite dimensional normed space, since it can be proved that the non empty weak open sets always contain a line.
Now, a way to prove the result:
I want to show that given a sequence $(x_n)\subset X$ and $x\in X$ such that $x_n\to x$ weakly, then $x_n\to x$ in the norm topology. 
Let $\{e_1,\dots,e_n \}$ be a basis of the vector space and denote by $\{e_1^*,\dots,e_n^* \}$ its dual basis. The convergence of the sequence in weak topology says in particular that $e^*_i(x_n)\to e^*_i(x)$ for all $i$. Then:
$$x_n= \sum_{i=1}^n e^*_i(x_n)\to \sum_{i=1}^n e^*_i(x) = x,$$
where the last convergence is in the norm topology. 
