Yes, your proof is correct. Here, I will just reword it to slightly improve clarity and precision .
Let $V$ be a vector space over $\mathbb R$ (or $\mathbb C$) with $\dim(V) = n$ and norm $\|\cdot\|$. Let $\{e_i\}_{i=1,\cdots , n}$ be a base of $V$. Suppose $v_k$ be a Cauchy sequence w.r.t. $\|\cdot\|$.
Since any two norms on a finite dimensional space are equivalent, $\|\cdot\|$ is equivalent to the $l^1$-norm $\|\cdot\|_1$. So, there are $C,D>0$ such that, for all $w\in V$, $C \|w\|_1 \leq \|w\| \leq D \|w\|_1$.
So, we have, for all $\varepsilon > 0$, there is $N$ such that, if $k,j>N$,
$$ \varepsilon > \|v_j - v_k\| \geq C \|v_j - v_k\|_1 = C \sum_{i=1}^n |v_{ji} - v_{ki}| \geq C |v_{ji} - v_{ki}|$$
for each $1 \leq i \leq n$. Hence $v_{ki}$ is a Cauchy sequence in $\mathbb R$ (or $\mathbb C$) for each $i$. Since $\mathbb R$ (or $\mathbb C$) is complete, there is $u_i$ in $\mathbb R$ (or $\mathbb C$) such that $u_i = \lim_{k \to \infty} v_{ki} $, for each $i$. Let $u = (u_1, \dots , u_n) = \sum_i u_i e_i$. Then, it is clear that, $u$ is in $V$.
Let us prove $\lim_{k \to +\infty} \|v_k - u\| = 0$:
$$ \lim_{k \to +\infty} \|v_k - u\| \leq D \lim_{k \to +\infty} \|v_k - u\|_1 = D \lim_{k \to +\infty} \sum_{i=1}^n |v_{ki} - u_i| = D \sum_{i=1}^n \lim_{k \to +\infty} |v_{ki} - u_i|=0$$
The main reason is the Poincaré inequality, the proof of which you can find, for example, here:
$$
\int_\Omega u^2 \le C\int_\Omega |\nabla u|^2 \quad \text{ provided } \int_\Omega u = 0
$$
The space $H^1(\Omega)$ is the direct sum of the one-dimensional space of constant functions, called $L$, and its orthogonal complement $M$, which consists of the functions with zero integral. By the Poincaré inequality, the norms are comparable on $M$. But the assumption (3), they are also comparable on $L$. This implies they are comparable on the entire space (all choices of norms on the direct sum of two normed spaces are equivalent.)
Best Answer
Yes this is true. If $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent, then by definition there exist $C > 0, c > 0$ such that $$c\|x\|_1 \le \|x\|_2 \le C\|x\|_1$$ for all $x \in V$.
Suppose that $(x_n) \subset V$ is Cauchy in the norm $\|\cdot\|_2$ and let $\epsilon > 0$. Then there exists an $N > 0$ such that $$\|x_n - x_m\|_2 \le c\epsilon$$ whenever $n,m > N$. Thus $$\|x_n - x_m\|_1 \le \epsilon$$ whenever $n,m > N$. So $(x_n)$ is Cauchy in the norm $\|\cdot\|_1$. The proof of the reverse implication works the same way.
Basically, equivalent norms induce the same open sets, so that as topological spaces $(V, \|\cdot\|_1)$ and $(V, \|\cdot\|_2)$ are the same. In other words, they induce equivalent metrics $d_1$ and $d_2$ on $V$ so that elements of $V$ are "close" with respect to $d_1$ iff they are "close" with respect to $d_2$.