The basic Cohen set has no countably infinite subset

Let $p⊩\dot{f}:\check{\omega }\to \dot{A}$, and enlarge a finite support $E$ of $\dot{f}$ to support $p$. If some $q\le p$ forces $\dot{f}(i)={\dot{a}}_n$ with $n\mathrm{\notin }E$, choose $m$ outside $E\cup \{n\}$ and outside the first-coordinate support of $q$. Let $\pi$ swap $n,m$. Then $\pi \dot{f}=\dot{f}$, $\pi p=p$, and F2 gives $\pi q⊩\dot{f}(i)={\dot{a}}_m$.

The conditions $q$ and $\pi q$ agree wherever both are defined: the $m$ coordinate was fresh and all other moved coordinates are fixed. Their union is a common extension forcing ${\dot{a}}_n={\dot{a}}_m$, contradicting F1. Hence no extension of $p$ can put a value outside $\{a_n:n\in E\}$, and density of decisions yields $p⊩\mathrm{ran}\dot{f}\subseteq \{{\dot{a}}_n:n\in E\}$.

Thus every such map has finite range, so none is injective or enumerates the infinite set $A$. A countably infinite subset would, by its meaning in ZF, carry a bijection from $\omega$ and hence an injection into $A$, also impossible. Fresh indices were chosen only from explicit complements of finite sets; no AC is used.