The Solovay model has no Hamel basis and no discontinuous additive real function

Suppose $H$ is a Hamel basis. It is nonempty because it spans $1$; choose one $b\in H$ (one existential choice, not AC). Define $c_b(x)$ as the unique rational coefficient of $b$ in the finite expansion of $x$. Then $c_b$ is additive, and F1 makes $W=\ker c_b$ a measurable proper subgroup. Moreover, $\mathbb{R}={\bigcup }_{q\in \mathbb{Q}}(qb+W)$. By F6, Countable Choice holds in $M$. If $\lambda (W)>0$, F3 gives $W=\mathbb{R}$; if $\lambda (W)=0$, translation invariance in F4 makes every $qb+W$ measurable and null, and countable subadditivity makes their explicitly rational-indexed union null. Monotonicity then gives $\lambda ([0,1])=0$, contradicting the value $1$ from F4.

Let $f$ be additive and put $E_n=\{x\in [-1,1]:\mathrm{\mid }f(x)\mathrm{\mid }\le n\}$. F1 makes these sets measurable, and they cover $[-1,1]$. By F6, Countable Choice holds in $M$. If each were null, F4 would make their explicitly indexed union null, contrary to $\lambda ([-1,1])=2$; hence some $E_n$ has positive measure. Steinhaus gives an interval about zero in $E_n-E_n$, where additivity bounds $\mathrm{\mid }f\mathrm{\mid }$ by $2n$. F5 then yields continuity and $f(x)=xf(1)$ for every real $x$. The zero map and $n=0$ cause no exception.

Step 1.1 excludes a basis, and step 1.2 excludes every discontinuous additive solution, without invoking an AC basis-existence theorem.