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 $\lim_{k \to +\infty} \|v_k - u\| = 0$:
$$ \lim_{k \to +\infty} \|v_k - u\| \leq D \lim_{k \to +\infty} \|v_k - u\|_1 = D \lim_{k \to +\infty} \sum_{i=1}^n |v_{ki} - u_i| = D \sum_{i=1}^n \lim_{k \to +\infty} |v_{ki} - u_i|=0$$
If $A$ has the norm $N$ and $N'$ is an equivalent norm, $N'$ is continuous. Since $N'(x)\neq 0$ for all $x\in A, f$ is continuous as well. A similar argument can be used to show the inverse is continuous.
There is an easier way however. Since $E$ is finite-dimensional, there is an isomorphism $L:E\rightarrow \mathbb{R}^n$ for some $n$. If $N$ is the norm on $E$, we can define a norm $N'$ on $\mathbb{R}^n$ by $N'(x):=N(L^{-1}(x))$ (you have to check this is a norm). Then $B=\{x\in \mathbb{R}^n:N'(x)=1\}$ is closed and bounded, since all norms on $\mathbb{R}^n$ are equivalent, hence B is compact. But then so is $A$, since $A=L^{-1}(B)$ and $L$ is an isomorphism.
Best Answer
It's just about showing that every sequence (bounded in the $||.||_\infty$ norm) $\{v_n\}\subset S_\infty$ of vectors has a convergent subsequence. For $w_n:=\phi(v_n)$, the corresponding sequence $\{ w_n \} $ has, by the virtue of compactness of $\phi(S_\infty)$, a convergent subsequence $\{w_{n_k}\}$ converging to some $w \in \phi(S_\infty)$. Now there is an obvious canditate for a subsequence of $\{v_n\}$ converging to $v := \phi^{-1}(w)$ and to show convergence one just needs to use the fact that $\phi$ is a linear isometry. This means that for any pair $x,y \in E $ of vectors, we have $||x-y|| = ||\phi (x-y)||=||\phi (x) -\phi (y)||$.