Proof of "big norm" lemma from Kreyszig section 2.4

Question

Problem statement:

I believe I have an alternate proof of Lemma 2.4-1 in Kreyszig's Introductory Functional Analysis with Applications which does not make use of Countable Choice, and I would like a second opinion on the validity of my work. The lemma is as follows:

2.4-1 Lemma (linear combinations) Let $\{x_1, \cdots, x_n\}$ be a linearly independent set of vectors in a (real or complex) normed space $(X,\|\cdot\|)$ (of any dimension). Then there is a number $c > 0$ such that for every choice of scalars $\alpha_1, \cdots, \alpha_n$ we have $$\|\alpha_1x_1 + \cdots + \alpha_nx_n\| \geq c(|\alpha_1| + \cdots + |\alpha_n|).$$

Kreyszig invokes Countable Choice in the proof by asserting (by way of contradiction) that there exists a sequence $(y_m) \subseteq \text{span}(\{x_1, \cdots, x_n\})$ which converges to zero in norm and yet whose coordinates have 1-norm equal to 1.

My attempt at a choice-free proof:

Preliminaries:

Let $Y = \text{span}(\{x_1, \cdots, x_n\})$, and also let $K$ ($= \mathbb{R}$ or $\mathbb{C}$) be the underlying field. Consider the map $L : K^n \to Y$ given by $$L(\alpha_1, \cdots, \alpha_n) = \alpha_1x_1 + \cdots + \alpha_nx_n.$$ Clearly $L$ is linear and surjective. Since the $x_i$ are linearly independent, $L$ is also injective. Proving the lemma is therefore equivalent to showing that there exists some $c > 0$ such that for all $y \in Y$, $$\|y\| \geq c\|L^{-1}(y)\|_1,$$ that is, $L^{-1}$ is bounded as a linear operator (here $\|\cdot\|_1$ denotes the 1-norm on $K^n$). We do this via four steps:

1. Prove $L$ is a bounded linear operator.
2. Show $Y$ is separable.
3. Find a suitable $c$ for the countable dense subset.
4. Show this same $c$ works for all $y \in Y$.

In fact these also show $L$ is a linear homeomorphism.

Step 1:

Let $(\alpha_1, \cdots, \alpha_n) \in K^n$ be given. If we let $M = \max\{\|x_1\|, \cdots, \|x_n\|\}$, then by norm properties, \begin{align*} \|L(\alpha_1, \cdots, \alpha_n)\| & = \|\alpha_1x_1 + \cdots + \alpha_nx_n\| \leq \|\alpha_1x_1\| + \cdots + \|\alpha_nx_n\| = |\alpha_1|\|x_1\| + \cdots + |\alpha_n|\|x_n\| \leq M(|\alpha_1| + \cdots + |\alpha_n|)\\ & \leq M\|(\alpha_1, \cdots, \alpha_n)\|_1. \end{align*} Hence $L$ is bounded, and therefore continuous.

Step 2:

Let $Q \subseteq K^n$ be a countable dense subset (e.g. $\mathbb{Q}^n$ in the real case, and $\mathbb{Q}[x]^n$ in the complex case). By continuous bijectivity of $L$, $$Y = L(K^n) = L\left(\overline{Q}\right) \subseteq \overline{L(Q)} \subseteq Y.$$ Therefore $\overline{L(Q)} = Y$, that is, $L(Q)$ is dense in $Y$. Since $Q$ is countable, we conclude $Y$ is separable.

Step 3:

Notice $L(Q)$ is countably infinite--write its elements as a sequence, e.g. $L(Q) = \{q_m : m \in \mathbb{Z}^+\}$. Suppose for contradiction that there is no $c > 0$ such that $\|q_m\| \geq c\|L^{-1}(q_m)\|_1$ for every $m \in \mathbb{Z}^+$. Then, for every $k \in \mathbb{Z}^+$, there exists an $m \in \mathbb{Z}^+$ such that $\|q_m\| < \frac{1}{k}\|L^{-1}(q_m)\|_1$.

Define a sequence $(m_k) \subseteq \mathbb{Z}^+$ by $$m_k = \min\left\{m \in \mathbb{Z}^+ : \|q_m\| < \frac{1}{k}\|L^{-1}(q_m)\|_1\right\}.$$ Necessarily $L^{-1}(q_{m_k})$ is nonzero by definition of $(m_k)$, because $\|q_{m_k}\| \geq 0$. Thus we may define a new sequence $(z_k) \subseteq Y$ by $$z_k = \frac{1}{\|L^{-1}(q_{m_k})\|_1}q_{m_k}.$$ Due to the above work, we see that $(\|z_k\|) \to 0$ and also $\|L^{-1}(z_k)\|_1 = 1$ for all $k$.

Repeat Kreyszig's argument in his proof of the lemma to conclude that $(z_k)$ has a convergent subsequence $(z_{k_s}) \to z \in Y$ such that $\|L^{-1}(z)\|_1 = 1$. This part of his proof does not require Countable Choice.

We must have $(\|z_{k_s}\|) \to \|z\|$ by continuity of the norm, yet $(\|z_{k_s}\|) \to 0$. Therefore $\|z\| = 0$ and hence $z = 0$. But the condition $\|L^{-1}(z)\|_1 = 1$ means that not all of the coordinates of $z$ are zero. Therefore $z \neq 0$ by linear independence of the $x_i$; this is a contradiction.

In conclusion, there must be a $c > 0$ so that $\|q\| \geq c\|L^{-1}(q)\|_1$ for all $q \in L(Q)$.

Step 4:

Let $y \in Y$ be given. By separability there exists (choice-free) a sequence $(q_k) \subseteq L(Q)$ such that $(q_k) \to y$. Because the set $\{q_k - y : k \in \mathbb{Z}^+\} \subseteq Y$ is countable, we may repeat the argument of Step 3 (or a shorter argument if the set is finite) to conclude there exists a $c_y > 0$ such that $$\|q_k - y\| \geq c_y\|L^{-1}(q_k - y)\|_1$$ for all $k$.

We claim $(L^{-1}(q_k)) \to L^{-1}(y)$ in $K^n$. Let $\epsilon > 0$ be given. Since $(q_k) \to y$, there exists an $N \in \mathbb{Z}^+$ such that $k \geq N$ implies $\|q_k - y\| < c_y \epsilon$. For such $k$, $$\|L^{-1}(q_k) - L^{-1}(y)\|_1 = \|L^{-1}(q_k - y)\|_1 \leq \frac{1}{c_y}\|q_k - y\| < \epsilon.$$ So indeed, $(L^{-1}(q_k)) \to L^{-1}(y)$ as desired.

Again by continuity of norm, we have $(\|L^{-1}(q_k)\|_1) \to \|L^{-1}(y)\|_1$ and $(\|q_k\|) \to \|y\|$. Because $\|q_k\| \geq c \|L^{-1}(q_k)\|_1$ for all $k$ by Step 3, we conclude that $\|y\| \geq c \|L^{-1}(y)\|_1$ and we're done.

Last remarks:

I used a few facts about continuity and convergence of sequences in $\mathbb{R}^n$ and $\mathbb{C}^n$, but none of them require Countable Choice. I would say it is somewhat important that this lemma be true without that axiom, since it is needed to prove that finite-dimensional normed spaces are Banach and all norms on such are equivalent. But of course, Kreyszig's original proof with Countable Choice is much quicker and more concise.

Thank you for any feedback!