This question is inspired by Pace Nielsen's recent question Does a left basis imply a right basis, without AC?.
For any field $k$, the field $k(x)$ of rational functions in one variable has an explicit $k$-basis given by partial fractions: the set
$$B(x,k)=\left\{x^i:i\geq 0\right\}\cup \left\{\frac{x^i}{m(x)^j}:\mbox{$m(x)$ monic irreducible, } 0<j, 0\leq i<\operatorname{deg}(m)\right\}$$
is a basis.
By induction, $k(x_1,\dots,x_n)$ has a $k$-basis for any $n$, without needing to assume the axiom of choice. The obvious way of doing this gives a basis that depends on the choice of an order for the variables.
In ZF set theory without choice, must $k(X)$ have a $k$-basis for an arbitrary set $X$?
The existence of such a basis certainly doesn't need the full strength of AC. The statement that every set has a total order is known to be independent of ZF but not to imply AC (see for example the various references in Are all sets totally ordered ?). If $X$ has a total order, and for $y\in X$ we let $$X_{<y}=\left\{x\in X: x<y\right\},$$
then the set of finite products $t_1t_2\dots t_n$, where
$$1\neq t_i\in B\left(y_i,k(X_{<y_i})\right)$$
for some $y_1<\dots < y_n$,
is a $k$-basis of $k(X)$.
Best Answer
I will show that, if $k$ is countable and has characteristic $0$, then $k(X)$ has a basis. (This whole proof takes place without choice.)
For any $x \in X$, and any $f \in X$, we define an element $r_x(f)$ of $k(X \setminus \{ x \})$ as follows: We have $k(X) \cong k(X \setminus \{ x \})(x)$, which embeds into the field of formal Laurent series $k(X \setminus \{ x \})((x))$. Let $r_x(f)$ be the coefficient of $x^0$ in this Laurent series. For a finite subset $I$ of $X$, let $r(I)$ be $$\{ f \in k(I) : r_x(f) =0\ \forall_{x \in I} \}.$$
For any finite set $I$, we have $k(I) = \bigoplus_{J \subseteq I} r(J)$ so, since $k(X)$ is the union of the $k(I)$ for $I$ finite, we have $k(X) = \bigoplus_{I \subseteq X,\ \# I < \infty} r(I)$. So it is enough to show $\bigoplus_{I \subseteq X,\ \# I < \infty} r(I)$ has a basis. For this purpose, it is enough to show that $r(\{ x_1, x_2, \ldots, x_n \})$ has an $S_n$ invariant basis. One can then transport this basis to each of the $r(I)$ with $\#I = n$. (More precisely, look at all $n!$ bijections between $\{ 1, \ldots, n \}$; each of them gives a transported basis in $r(I)$. Using the $S_n$ invariance, their union is also a basis in $r(I)$. Thus, we get a basis in $r(I)$ without having to choose an ordering of $I$.)
We fix particular representatives $Sp_{\lambda}$ for the irreducible representations of $S_n$. This can be done in a choice free way by, for example, writing down formulas with Young tableaux. For any $S_n$-representation $W$, let $W_{\lambda} = \mathrm{Hom}_{S_n}(Sp_{\lambda}, W)$. Then there is an obvious map $\bigoplus_{|\lambda| = n} W_{\lambda} \otimes Sp_{\lambda} \longrightarrow W$ and representation theory tells us this map is an isomorphism. We will show that $r(\{ x_1, \ldots, x_n \})_{\lambda}$ has a countably infinite basis for each $\lambda$, and we will pick a particular one without choice. Thus $$r(\{x_1, \ldots, x_n \}) \cong \bigoplus_{|\lambda| = n} k^{\oplus \omega} \otimes Sp_{\lambda} \cong (k S_n)^{\bigoplus \omega}$$ as $S_n$ representations, and the latter has an obvious permutation basis.
Lemma: Any subspace $V$ of $k^{\oplus \omega}$ has a countable or finite basis, which can be found with no choices.
Proof: Let $e_i$ be the standard basis vectors of $k^{\oplus \omega}$. Let $k^n$ be the subspace of $k^{\oplus \omega}$ spanned by $e_1$, $\dots$, $e_n$. Let $S \subseteq \omega$ be the set of $n$ for which $V \cap k^n \neq V \cap k^{n-1}$. For each $s \in S$, there is a unique $v_s \in V$ such that $v_s \in k^n \cap V$, the coefficient of $e_s$ in $v_s$ is $1$, and $v_s$ does not contain the basis vectors $e_{s'}$ for $s' \in S$, $s < n$. These $v_s$ form a basis. (In other words, we use row reduced echelon form.) $\square$
Now, $k(x_1, \ldots, x_n)$ has a countable basis. So $r(\{ x_1, \ldots, x_n \})$ has a countable or finite basis by the lemma. The Specht module $Sp_{\lambda}$ has an explicit finite basis which can be written down using standard Young tableaux, so $\mathrm{Hom}(Sp_{\lambda}, r(\{ x_1, \ldots, x_n \}))$ has a basis (note that this is $\mathrm{Hom}$ of vector spaces). The $S_n$ equivariant maps are a subspace of this so, using the lemma again, $\mathrm{Hom}_{S_n}(Sp_{\lambda}, r(\{ x_1, \ldots, x_n \}))$ has an explicit finite or countable basis, as desired.
Finally, we must check that $\mathrm{Hom}_{S_n}(Sp_{\lambda}, r(\{ x_1, \ldots, x_n \}))$ is infinite dimensional. The easiest way is probably to note that, for any distinct positive integers $(a_1, \ldots, a_n)$, the span of $x_1^{a_{\sigma(1)}} \cdots x_n^{a_{\sigma(n)}}$ forms a copy of $k S_n$ inside $r(\{ x_1, \ldots, x_n \})$, and these are all linearly independent.