While the question is still unanswered, let me make a couple of observations, showing that the problem is essentially equivalent to construction of symmetric bases (this is already alluded to in Is there a symmetric basis for $\mathbf{Q}(x,y)$? , however we really need infinitely many variables).

Any permutation $\pi$ of $X$ extends uniquely to a $k$-automorphism of $k(X)$, which I will also denote $\pi$. Let me say that a basis $B$ of $k(X)$ is *symmetric* if $\pi(B)=B$ for every permutation $\pi$, and *almost symmetric* if there is a finite set $X_0\subseteq X$ such that $\pi(B)=B$ for every permutation $\pi$ fixing $X_0$ pointwise.

**Proposition 1:** ZF proves that if $k(\{x_n:n\in\mathbb N\})$ has an almost symmetric basis, then $k(X)$ has a basis for every set of variables $X$.

**Proof:** Pick a basis $B$ of $k(\{y_n:n\in\mathbb N\})$ invariant under all permutations fixing $Y_0=\{y_1,\dots,y_m\}$. If $X$ is finite, $k(X)$ has a basis by induction on $|X|$, for example the one in the question, hence we can assume $X$ is infinite. Fix distinct elements $x_1,\dots,x_m\in X$, and put
$$B_X=\{f(x_1,\dots,x_n):n\ge m,f(y_1,\dots,y_n)\in B,x_{m+1},\dots,x_n\in X\}$$
where the $x_{m+1},\dots,x_n$ are pairwise distinct and disjoint from $\{x_1,\dots,x_m\}$ (this did not fit into the displayed line).

Since any element of $k(X)$ is contained in $k(x_1,\dots,x_n)$ for some $x_1,\dots,x_n$ as above, $B_X$ generates $k(X)$. On the other hand, any finite subset $B_0\subseteq B'$ is also included in some $k(x_1,\dots,x_n)$, and then the symmetry of $B$ implies that $\{f(y_1,\dots,y_n):f(x_1,\dots,x_n)\in B_0\}\subseteq B$, hence it is linearly independent.

**Proposition 2:** Conversely, if ZF proves that every $k(X)$ has a basis, then ZF(C?) proves that $k(\{x_n:n\in\mathbb N\})$ has an almost symmetric basis.

**Proof:** Assume (in a model of ZF) that $k(\{x_n:n\in\mathbb N\})$ has no almost symmetric basis, we will extend the universe into a permutation model of ZFA where some $k(X)$ has no basis. This can be made into a model of ZF by the Jech–Sochor embedding theorem; however, I suspect that the forcing involved in that step may need some choice in the ground model, which is why I put the half-hearted C in the statement.

As for the permutation model: we extend the universe with a countable set of atoms $A=\{a_n:n\in\mathbb N\}$. We take the group of all permutations of $A$, and the normal filter generated by stabilizers of finite subsets of $A$. In the resulting permutation model, $k(A)$ has no basis $B$: otherwise $B$ would also be a basis of $k(A)\simeq k(\{x_n:n\in\mathbb N\})$ in the ground model, and it would have finite support, which is exactly what I called almost symmetric above.

**Proposition 3:** ZF proves that $k(x_1,\dots,x_n)$ has an explicitly definable symmetric basis for every field $k$ and $n\in\mathbb N$.

**Proof:** Let $B_0$ be any basis of $k(y_1,\dots,y_n)$, which exists by induction on $n$ (e.g., we can take the one described in the question). Let $s_k(x_1,\dots,x_n)$ denote the $k$th elementary symmetric polynomial, and put
$$K=k(s_1(x_1,\dots,x_n),\dots,s_n(x_1,\dots,x_n))\subseteq k(x_1,\dots,x_n).$$
Since the $s_k(\vec x)$ are algebraically independent, $K$ is isomorphic to $k(y_1,\dots,y_n)$, and
$$B_1=\{f(s_1(x_1,\dots,x_n),\dots,s_n(x_1,\dots,x_n)):f(y_1,\dots,y_n)\in B_0\}$$
is a basis of $K$ over $k$. The field $L=k(x_1,\dots,x_n)$ is the splitting field of the degree $n$ polynomial
$$h(x)=(x-x_1)\cdots(x-x_n)$$
over $K$, hence $[L:K]\le n!$. (Actually, it is easy to see that $\mathrm{Gal}(L/K)=S_n$, hence the degree is exactly $n!$.)

I claim that
$$B_2=\{x_1^{\pi(1)}\cdots x_n^{\pi(n)}:\pi\in S_n\}$$
is a $K$-basis of $L$, hence $B=B_1B_2$ is a $k$-basis of $L$, and it is obviously symmetric.

Since we already know $|B_2|\ge[L:K]$, it suffices to show that $B_2$ is $K$-linearly independent. I have no high-level argument for that, but one can prove it by induction on $n$. It will be more convenient to label the variables as $x_0,\dots,x_{n-1}$, and consider $S_n$ as permutations on $\{0,\dots,n-1\}$.

The case $n=1$ is trivial, so assume the statement holds for $n$, we will show it for $n+1$. Aiming for contradiction, let
$$\sum_{\pi\in S_{n+1}}f_\pi(x_0,\dots,x_n)\prod_{i=0}^nx_i^{\pi(i)}=0,\tag{$*$}$$
where the $f_\pi\in K$ are not all zero. We may assume without loss of generality that the rational functions $f_\pi$ are in fact polynomials, and that $x_0\cdots x_n$ does not divide all $f_\pi$.

Let $i_0\le n$. If we substitute $0$ for $x_{i_0}$ in $(*)$, and rename $x_{i_0+1},\dots,x_n$ to $x_{i_0},\dots,x_{n-1}$, we get
$$x_0\cdots x_{n-1}\sum_{\pi'\in S_n}f_{\pi}(x_0,\dots,x_{i_0-1},0,x_{i_0},\dots,x_{n-1})\prod_{i=0}^{n-1}x_i^{\pi'(i)}=0,$$
where
$$\pi(i)=\begin{cases}
\pi'(i)+1&i<i_0,\\
0&i=i_0,\\
\pi'(i-1)+1&i>i_0.
\end{cases}$$
By the induction hypothesis, we obtain that $f_\pi(x_0,\dots,x_{i_0-1},0,x_{i_0+1},\dots,x_n)=0$ whenever $\pi(i_0)=0$. Thus, any $f_\pi$ is divisible by $x_{\pi^{-1}(0)}$, and as it is symmetric, by $x_0\cdots x_n$. This contradicts one of our assumptions.

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