A Banach space has no countably infinite Hamel basis

Let $Q_{\mathbb{K}}:=\mathbb{Q}$ in the real case and $Q_{\mathbb{K}}:=\{a+ib:a,b\in \mathbb{Q}\}$ in the complex case. In either case $Q_{\mathbb{K}}$ is countable by [L3], and it is dense in $\mathbb{K}$. Let $D\subseteq X$ be the set of all finite $Q_{\mathbb{K}}$-linear combinations of the basis vectors $b_n$. Coding a finite combination by the finite list of its indices together with its coefficient list gives an explicit surjection from a countable set onto $D$, so $D$ is countable. Also $0\in D$, so $D$ is nonempty.

For $N\in \mathbb{N}$ put $F_N:={\mathrm{span}}_{\mathbb{R}}\{b_0,\dots ,b_N\}$ in the real case, and $F_N:={\mathrm{span}}_{\mathbb{R}}\{b_0,i{b}_0,\dots ,b_N,i{b}_N\}$ in the complex case. In either case $F_N$ is finite-dimensional over $\mathbb{R}$. It is proper: in the real case $b_{N+1}\mathrm{\notin }{F}_N$ by linear independence of the basis image, and in the complex case a real-linear relation expressing $b_{N+1}$ in terms of $b_0,i{b}_0,\dots ,b_N,i{b}_N$ would be the same as a complex-linear relation expressing $b_{N+1}$ in terms of $b_0,\dots ,b_N$. Thus [L2] makes every $F_N$ closed. Also $X={\bigcup }_{N\in \mathbb{N}}{F}_N$, because every vector uses only finitely many basis vectors, and in the complex case every complex coefficient splits into real and imaginary parts.

Every proper linear subspace of a normed space has empty interior. Indeed, if $W⊊X$ were a linear subspace containing some ball $B(x,r)$, then $B(0,r)\subseteq W$ because $W$ is closed under subtraction, and for any $y\in X\setminus \{0\}$ the vector $(r\mathrm{/}(2\mathrm{\parallel }y\mathrm{\parallel }))y$ would lie in $B(0,r)\subseteq W$, forcing $y=(2\mathrm{\parallel }y\mathrm{\parallel }\mathrm{/}r)(r\mathrm{/}(2\mathrm{\parallel }y\mathrm{\parallel }))y\in W$; also $0\in W$. So $W=X$, contradiction.

By [L5], fix a surjection $d:\mathbb{N}\to D$.

$D$ is dense in $X$. Indeed, let $x={\sum }_{j=0}^m{\lambda }_j{b}_{n_j}\in X$ and let $\epsilon >0$. Choose $q_j\in Q_{\mathbb{K}}$ with $\mathrm{\mid }{\lambda }_j-q_j\mathrm{\mid }<\epsilon \mathrm{/}(2(m+1)(1+{\sum }_{j=0}^m\mathrm{\parallel }{b}_{n_j}\mathrm{\parallel }))$. Then $\parallel x-{\sum }_{j=0}^m{q}_j{b}_{n_j}\parallel \le {\sum }_{j=0}^m\mathrm{\mid }{\lambda }_j-q_j\mathrm{\mid }\mathrm{\parallel }{b}_{n_j}\mathrm{\parallel }<\epsilon \mathrm{.}$ So every vector of $X$ lies in the closure of $D$.

We now run the separable-complete Baire argument inside the open unit ball $U_0:=\{x:\mathrm{\parallel }x\mathrm{\parallel }<1\}$. Because $F_0$ is closed with empty interior, the set $U_0\setminus F_0$ is nonempty and open. By step 2.2, the set $A_0:=\{m\in \mathbb{N}:d(m)\in U_0\setminus F_0\}$ is nonempty, so [L5] gives its least element $m_0$. Put $x_0:=d(m_0)$. Since $U_0\setminus F_0$ is open at $x_0$, the set $B_0:=\{k\in {\mathbb{N}}_{\ge 1}:\bar{B}(x_0,\frac{1}{k})\subseteq U_0\setminus F_0\}$ is nonempty; let $k_0$ be its least element and set $r_0:=1\mathrm{/}{k}_0$. Then $\bar{B}(x_0,r_0)\subseteq U_0\setminus F_0$.

Inductively, if closed balls $\bar{B}(x_n,r_n)\subseteq U_0\setminus F_n$ have been chosen with $\bar{B}(x_n,r_n)\subseteq B(x_{n-1},r_{n-1}\mathrm{/}2)$ for $n\ge 1$, then $B(x_n,r_n\mathrm{/}2)\setminus F_{n+1}$ is a nonempty open set. By step 2.2, the set $A_{n+1}:=\{m\in \mathbb{N}:d(m)\in B(x_n,r_n\mathrm{/}2)\setminus F_{n+1}\}$ is nonempty, so [L5] gives its least element $m_{n+1}$. Put $x_{n+1}:=d(m_{n+1})$. Since $B(x_n,r_n\mathrm{/}2)\setminus F_{n+1}$ is open at $x_{n+1}$, the set $B_{n+1}:=\{k\in {\mathbb{N}}_{\ge 1}:\frac{1}{k}<\frac{r_n}{2}\text{ and }\bar{B}(x_{n+1},\frac{1}{k})\subseteq B(x_n,r_n\mathrm{/}2)\setminus F_{n+1}\}$ is nonempty; let $k_{n+1}$ be its least element and set $r_{n+1}:=1\mathrm{/}{k}_{n+1}$. Then $\bar{B}(x_{n+1},r_{n+1})\subseteq B(x_n,r_n\mathrm{/}2)\setminus F_{n+1}\subseteq \bar{B}(x_n,r_n),$ and $r_{n+1}<r_n\mathrm{/}2$. Hence $r_n\le r_0\mathrm{/}{2}^n$ for every $n$, so $r_n\to 0$.

For $m>n$, the inclusion from step 3.2 gives $x_m\in B(x_n,r_n\mathrm{/}2)$, so $\mathrm{\parallel }{x}_m-x_n\mathrm{\parallel }<r_n\mathrm{/}2$. Hence $(x_n)$ is Cauchy. Since $X$ is Banach, [L1] gives $x_n\to x$ for some $x\in X$. Each $\bar{B}(x_n,r_n)$ is closed and contains all later $x_m$, so it contains the limit $x$; therefore $x\in \bar{B}(x_n,r_n)\subseteq X\setminus F_n$ for every $n$.

Step 4.1 contradicts $X={\bigcup }_N{F}_N$ from step 1.2. Therefore no countably infinite Hamel basis exists. The foundational point recorded in [L4] is that the proof used only a fixed countable dense set, with both the recurring point selections and the ball radii chosen canonically from $\mathbb{N}$, and not the unrestricted complete-metric Baire theorem.