There are a lot of questions here on showing that two norms are not equivalent. I understand that two norms may not be equivaelent from their proofs, however I do not understand why this happened in the first place. Isn't there a theorem which says that: If we are given two norms on some finite-dimensional vector space V over C, they are always within a constant factor of one another.
What am I missing ?
Norms on finite dimensional spaces are equivalent. Norms on infinite dimensional spaces are not always equivalent.
Let $\left \| \cdot \right \| _{a}$ and $\left \| \cdot \right \| _{b}$ be norms on $V$.
The key idea is that if $\dim V<\infty $ then the unit ball in the topology induced by both norms is compact.
This means in particular that there is an $M\geq 0$ such that $\left \| x \right \|_{a}\leq M$ for all $x\in V$ such that $\left \| x \right \|_{b}\leq1$.
But then $\left \| \frac{x}{\left \| x \right \|_{b}} \right \|_{a}\leq M$ for all $x\in V$, or what is the same thing $\left \| x \right \|_{a}\leq M\left \| x \right \|_{b}$. Reversing the roles of the norms, we get $\left \| x \right \|_{b}\leq M'\left \| x \right \|_{1}$ and the result follows.
Note: one does need to show that $\left \| \cdot \right \|_{a}:V\to \mathbb R$ is continuous on $V$ in the topology induced by $\left \| \cdot \right \|_{b}$ but this is fairly routine.
The theorem you're referring seems to be this one:
Two norms $\lVert \cdot \lVert_\alpha$ and $\lVert \cdot \lVert_\beta$ on a vector space $V$ over $\Bbb R$ are equivalent if and only if there exist positive real numbers $c$ and $d$ such that $c\lVert x \lVert_\alpha \leq \lVert x \lVert_\beta \leq d\lVert x \lVert_\alpha$ for all $x \in V$.
(Taken from https://en.wikipedia.org/wiki/Norm_(mathematics)#Properties)
You might also be interested in this question about the proof of that theorem: Understanding of the theorem that all norms are equivalent in finite dimensional vector spaces
