Equivalent norms without Cauchy-Schwarz inequality

Let $X$ be a finite-dimensional vector space over $\mathbb{F}$. ($\mathbb{R}$ or $\mathbb{C}$)

Theorem: All norms on $X$ are equivalent.

Proof: $a_k$s and $c_k$s will refer to elements of $\mathbb{F}$. Let $(e_k)_{k=0}^{n-1}$ be a basis of $X$. Define
$\lVert \sum_{k=0}^{n-1} c_k e_k \rVert_1 = \sum_{k=0}^{n-1} |c_k|$. Evidently $\lVert \cdot \rVert_1$ is a norm. Let $\lVert \cdot \rVert$ be another norm on $X$. Then if we define $c_0 = 1/\max(\lVert e_k \rVert)_{k=0}^{n-1}$, for $x = \sum_{k=0}^{n-1} a_k e_k$,
$$ c_0\lVert x \rVert = c_0\left\lVert \sum_{k=0}^{n-1} a_k e_k \right\rVert
\leq c_0 \sum_{k=0}^{n-1} |a_k| \lVert e_k \rVert \leq \sum_{k=0}^{n-1} |a_k| = \lVert x \rVert_1
$$
Conversely, observe that if $c_1 > 0$ so that $\lVert x \rVert_1 \leq c_1 \lVert x \rVert$, (We may assume that $\lVert x \rVert_1 \neq 0$ by the identity of indiscernibles)
$$ 1 = \frac{\lVert x \rVert_1}{\lVert x \rVert_1} \leq \frac{c}{\lVert x \rVert_1} \lVert x \rVert = c_1\left\lVert \frac{x}{\lVert x \rVert_1} \right\rVert
$$
so it suffices to show that $\lVert\cdot\rVert$ is bounded below on the unit sphere $S =\{x \in X\mid \lVert x \rVert_1 = 1\}$.

However, this is the part I’m having trouble with. Obviously if $\mathbb{F} = \mathbb{R}$, I can use the fact that every bounded set in $\mathbb{R}^n$ is totally bounded, and thus show the compactness of $S$.(which implies every sequence of norms on $S$ has a convergent subsequence) But what if $\mathbb{F} = \mathbb{C}$? How does the argument work here? Also, the sequence of norms mentioned earlier converges iff $\lVert\cdot\rVert$ is continuous, which is what I’m trying to show.

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The same argument work, because compact set in $\mathbb{C}^n$ are exactly closed and bounded set (think about $\mathbb{C}^n$ as $\mathbb{R}^{2n}$), but you can also avoid all this argument by using Open mapping theorem as above :

the identity maps :
$$
\begin{array}{c}
I : (E,\|.\|_1) \rightarrow (E, \|.\|)
\end{array}
$$
Is surjective and continuous (that what you have proved) and the space $E$ is a Banach space for $\|.\|_1$ so it inverse will be continuous.