Can someone verify that everything I did in this proof is correct?
Every two norms on a finite dimensional (real or complex) vector space $V$ are equivalent.
Proof: We assume that $\dim(V)=n$ and that $V$ is a real vector space. If $V$ is complex, the proof is completely analogue.
Let $E:= \{e_1, \dots, e_n\}$ be a basis of $V$. Write $v = \sum_i v_ie_i$ for every vector $v \in V$.
Then, define $\Vert\cdot\Vert_2: V \to \mathbb{R}^+: v \mapsto \left(\sum_{i}|v_i|^2\right)^{1/2}$. This is a norm on $V$. We prove that every norm is equivalent with this norm, and then the result will follow because the relation "two norms are equivalent" is an equivalence relation.
Let $\Vert \cdot\Vert$ be an arbitrary norm on $V$. Let $M:= (\sum\Vert e_i\Vert^2)^{1/2}$. Then, we have, for $v \in V:$
$$\Vert v\Vert = \Vert \sum v_ie_1 \Vert \leq \sum|v_i|\Vert e_i\Vert \leq (\sum|v_i|^2)^{1/2}(\sum\Vert e_i\Vert^2)^{1/2} = M\Vert v\Vert_2$$ where we used Cauchy's inequality for the last inequality.
It also follows that $\Vert\cdot \Vert: (V, \Vert \cdot \Vert_2) \to \mathbb{R}^+$ is a continuous function. Indeed, let $w \in V, \epsilon > 0$
Then, for $v \in V$ satisfying $\Vert v-w\Vert _2 < \delta$, it follows that:
$$|\Vert v\Vert - \Vert w\Vert| \leq \Vert v-w\Vert \leq M\Vert v - w \Vert_2 < M \delta = \epsilon$$
if we take $\delta :=\epsilon/M$.
Now, we prove that $S^{n-1}:= \{v \in V \mid \Vert v \Vert_2 =1\}$ is compact in $(V,\Vert \cdot \Vert_2)$. Because $V$ is finite dimensional, it suffices to prove that $S^{n-1}$ is closed and bounded. Boundedness is trivial.
For closedness, it suffices to notice that $S^{n-1}= \Vert \cdot \Vert_2^{-1}(\{1\})$ and because $\Vert \cdot \Vert_2: (V, \Vert \cdot\Vert_2) \to \mathbb{R}^+$ is continuous and $\{1\}$ is closed, it follows that $S^{n-1}$ is closed in $(V, \Vert \cdot \Vert_2)$
Hence, the function $\Vert \cdot \Vert: S^{n-1} \to \mathbb{R}^+$ attains a minimum $m > 0$ ($m=0$ is not possible because the domain are vectors with norm 1)
So, for every $v \in S^{n-1}$, we have:
$$0 < m \leq \Vert v \Vert$$
and hence, for $v \neq 0$, it follows that:
$$m \leq \Vert \frac{1}{\Vert v \Vert_2} v \Vert$$
and for $v=0$ the inequality is obvious.
From this, it follows that $m \Vert v\Vert _2 \leq \Vert v \Vert$ such that the two norms are equivalent, as desired.
I found nothing wrong with your proof, but perhaps that you could justify the assertion “Boundedness is trivial”.