Here's the argument that you want in the end.
Suppose you have an arbitrary norm on $\mathbb R^n$, call it $\| \cdot \|_{\alpha}$ (I don't know why you put $\mathbb n$ in the exponent here, that's not necessary). Write $\| \cdot \|_2$ for the standard Euclidean norm (the $2$-norm of the $L^p$-norms).
You need to be careful with topological arguments, especially when discussing normed spaces, because different norms induce different topologies. But for the sake of discussion, we only endow $\mathbb R^n$ with the topology coming from the $\| \cdot \|_2$ norm, so there's only one topology involved and Heine-Borel's theorem (compact $\Leftrightarrow$ closed and bounded) holds just fine as it's the classical topology in which it is usually proven.
So you've shown that $\| \cdot \|_{\alpha}$ is a continuous function on the unit sphere (that's your set $S$, namely $\{ x \in \mathbb R^n : \|x\|_2 = 1 \}$):
$$
|\|x\|_{\alpha} - \|y\|_{\alpha}| \le \|x-y\|_{\alpha} \le A \sqrt n \|x\|_2
$$
where $A = \max\{\|e_1\|_{\alpha}, \cdots, \|e_n\|_{\alpha}\}$ (kudos for seeing the argument there!). This shows that $\|x\|_{\alpha}$ is a $K$-Lipschitz function with $K = A \sqrt n$, and therefore is continuous. You used the sequential argument for that, but it's always cool to see the properties our functions have, just in case it's useful.
The function $x \mapsto \|x\|_{\alpha}$ is continuous on a compact, thus attains its minimum and maximum on $S$. Call them $m$ and $M$ respectively. Therefore,
$$
m \le \left\| \frac {x}{\left\| x \right\|_2} \right\|_{\alpha} \le M
$$
and so
$$
m \left\| x \right\|_2 \le \left\| x \right\|_{\alpha} \le M \left\| x \right\|_2.
$$
Your remaining question: why is $m > 0$? Well, $m = \|x\|_{\alpha}$ for $x \in S$. If $m=0$, then since $\|x\|_{\alpha}$ is a norm, that would imply $x = 0$, but $x \in S$, a contradiction.
Also note that norm equivalence is an equivalence: since we've proven that $\| \cdot \|_{\alpha} \sim \|\cdot \|_2$, we don't have to prove $\| \cdot \|_2 \sim \|\cdot \|_{\alpha}$; it immediately follows that
$$
(1/M) \left\| x \right\|_{\alpha} \le \left\| x \right\|_2 \le (1/m) \left\| x \right\|_{\alpha}.
$$
Edit: As requested in the comments, you wanted to compute $m$ and $M$ in the case where $\| \cdot \|_{\alpha}$ is the $1$-norm. If you follow the steps of the proof, we have to minimize/maximize the $1$-norm over the unit sphere (in the $2$-norm, so the standard unit sphere that looks like an actual sphere!).
So were looking to find extrema of $\sum_{i=1}^n |\lambda_i|$ on the surface $\sum_{i=1}^n \lambda_i^2 = 1$. By symmetry, we can assume all $\lambda_i \ge 0$. Since they are now positive, we can also replace $\lambda_i$ by their squares. So we now have a smooth problem: we try to find extremas of $f(\lambda_1, \cdots, \lambda_n) = \sum_{i=1}^n \lambda_i^2$ on the surface $g(\lambda_1,\cdots,\lambda_n) = \sum_{i=1}^n \lambda_i^4 = 1$. We still want $\lambda_i \ge 0$, but the same symmetry is present, so we don't need to worry about boundary conditions; we can literally ignore them. Note that even in the new problem, we'll still find the same extremal values (i.e. minimum and maximum value taken), although the inputs won't be the same, but we don't care about that right now since we are computing $m$ and $M$.
Now we compute the gradients of both functions $f$ and $g$:
$$
\nabla f(\lambda_1,\cdots,\lambda_n) = 2(\lambda_1,\cdots,\lambda_n), \qquad \nabla g(\lambda_1,\cdots,\lambda_n) = 4(\lambda_1^3, \cdots, \lambda_n^3).
$$
so we know that at extremal points, there exists $c$ with
$$
2(\lambda_1,\cdots,\lambda_n) = \nabla f = c \nabla g = 4c (\lambda_1^3, \cdots, \lambda_n^3)
$$
This now leads to the set of $n$ equations
$$
2c \lambda_i^3 - \lambda_i = 0.
$$
We cannot have all $\lambda_i$ equal to zero, so one of them is not. Say it is $\lambda_1$. This leads to
$$
2c \lambda_1^2 - 1 = 0 \quad \Rightarrow \quad \lambda_1 = \frac{\pm 1}{\sqrt{2c}}.
$$
This means that each $\lambda_i$ can only take the values $- \frac 1{\sqrt{2c}}$, $0$, or $\frac 1{\sqrt{2c}}$.
The sign doesn't affect the values of $f$, so we'll omit the negative case moving forward.
To compute $c$, take all the indices $i$ for which $\lambda_i \neq 0$, call $I$ the set of such $i$'s and notice that
$$
1 = \sum_{i=1}^n \lambda_i^4 = \sum_{i \in I} \lambda_i^4 = \sum_{i \in I} \left( \frac 1{\sqrt{2c}} \right)^4 = \frac{|I|}{4c^2}
$$
so that $c = \frac{\sqrt{|I|}}2$. For each value of $|I|$, we get the values of $f$ corresponding to potential extremas:
$$
\sum_{i=1}^{|I|} \lambda_i^2 = \sum_{i=1}^{|I|} \left(\frac 1{\sqrt {2c(|I|)}} \right)^2 = \sum_{i=1}^I \frac 1{2c(|I|)} = \frac{|I|}{2 \frac{\sqrt{|I|}}2 } = \sqrt{|I|}.
$$
Since $|I|$ takes values ranging from $1$ to $n$, we see that $m=1$ and $M=\sqrt n$.
Hope that helps,
Best Answer
Your proof of the triangle inequality is wrong, because you worked as if $d(x,y)=\|x-y\|_2$. You must consider each possibility: