This question is an extension of this one.
Let $F_X$ denote the free group on the set $X$. For any group $G$ and subset $S\!\subseteq\!G$, $\langle S\rangle$ denotes the subgroup generated by $S$ and $\mathrm{rank}(G) :=\min\{|S|;\:S\!\subseteq\!G, \langle S\rangle\!=\!G\}$.
PROPOSITION:
a) $F_X\cong F_Y\:\Leftrightarrow\:|X|=|Y|$
b) $\mathrm{rank}(F_X)=|X|$
Thus for every cardinal number $c$, there is (up to isomorphism) exactly one free group of rank $c$.
Proof:
a) $(\Leftarrow)$: If $f\!:X\rightarrow Y$ is the bijection, then $\varphi(x_1\ldots x_k):=f(x_1)\ldots f(x_k)$ is the isomorphism.
$(\Rightarrow)$: $F_X\!\cong\!F_Y$ $\Rightarrow$ $\mathrm{Ab} F_X\!\cong\!\mathrm{Ab} F_Y$ $\Rightarrow$ $\oplus_{x\in X}\mathbb{Z}\!\cong\!\oplus_{y\in Y}\mathbb{Z}$, so $|X|\!=\!|Y|$, since rank is known to be an invariant of free modules.
Alternatively, $\big(\oplus_{x\in X}\mathbb{Z}\big)\otimes_\mathbb{Z}\mathbb{Q}$ $\cong$ $\oplus_{x\in X}\big(\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Q}\big)$ $\cong$ $\oplus_{x\in X}\mathbb{Q}$, so $\oplus_{x\in X}\mathbb{Q}$ $\cong$ $\oplus_{y\in Y}\mathbb{Q}$, even as $\mathbb{Q}$-modules, but isomorphic vector spaces are known to have equipollent bases.
b) Since $\langle X\rangle\!=\!F_X$, $\mathrm{rank}(F_X)\leq|X|$. Suppose we have $Y\!\subseteq\!F_X$, $\langle Y\rangle\!=\!F_X$, $|Y|\!<\!|X|$.
QUESTION: how can I finish the proof of b), i.e. prove that $F_X$ can't be generated by a subset with smaller cardinality than $|X|$?