Although the question has an accepted answer, I will post a forcing-free proof of this fact that uses the Baire category theorem.

The theorem Joel proved is originally due to Harvey Friedman and is a special case of his "Borel diagonalization theorem" that he proves in his paper *On the necessary use of abstract set theory, Advances in Mathematics, 41 (1981), 209-280.* The following proof is from the appendix of this paper.

> **Theorem**. Assume that $f: {\mathbb{R}}^{\mathbb{N}} \rightarrow \mathbb{R}$ is a Borel measurable function with the property that if $x =^+ y$, then $f(x)=f(y)$, where $x =^+ y$ if and only if $\{x_n: n \in \mathbb{N}\}=\{y_n: n \in \mathbb{N}\}$ for $x,y \in {\mathbb{R}}^{\mathbb{N}}$. Then, there exists $x \in {\mathbb{R}}^{\mathbb{N}}$ such that $f(x)=x_k$ for some $k \in \mathbb{N}$.

**Proof**. Let $f$ be such a Borel map. In particular, $f(x)=f(y)$ whenever $g \cdot x = y$ for some $g \in Fin(\mathbb{N})$ where the action is given by permuting the indices.

For each rational $q \in \mathbb{Q}$, define $A_q=\{x \in {\mathbb{R}}^{\mathbb{N}}: f(x) < q\}$. Let ${\overline{\mathbb{R}}}^{\mathbb{N}}$ denote the product space where $\overline{\mathbb{R}}$ is the space of real numbers endowed with the discrete topology.

**Lemma.** For any $q \in \mathbb{Q}$, either $A_q$ or $A_q^C$ is meager in  ${\overline{\mathbb{R}}}^{\mathbb{N}}$.

**Proof**. Pick some $q \in \mathbb{Q}$. Then, $A_q$ is also Borel in ${\overline{\mathbb{R}}}^{\mathbb{N}}$ and hence has the Baire property. Let $U$ be an open set such that $A_q \triangle U$ is meager.

Each $g \in Fin(\mathbb{N})$ induces a homeomorphism of ${\overline{\mathbb{R}}}^{\mathbb{N}}$. Then, since $f$ is producing the same element for sequences that are in the same orbit under the action of $Fin(\mathbb{N})$, $g[A_q] \triangle g[U]=A_q \triangle g[U]$. So, each $A_q \triangle g[U]$ is meager and so is $A_q \triangle \bigcup_{g} g[U]$.

If $U=\emptyset$, then $A_q$ is meager. If $U \neq \emptyset$, then $\bigcup_{g} g[U]$ is dense and it follows that $A_q^C$ is meager since $(\bigcup_{g} g[U])^C$ and $\bigcup_{g} g[U] - A_q$ are meager. This completes the proof of the lemma. $\blacksquare$

Now, let $S$ be the set of all rationals $q$ such that $A_q$ is meager in ${\overline{\mathbb{R}}}^{\mathbb{N}}$. $S$ cannot be $\emptyset$ or $\mathbb{Q}$ by the Baire category theorem. Let $z=sup(S)$. Then, since $\bigcup_{s < z} A_q \cup \bigcup_{s > z} A_q^C$ is meager, $\{x \in {\mathbb{R}}^{\mathbb{N}}: f(x) = z\}$ is comeager, and hence dense by the BCT. But then, $\{x \in {\mathbb{R}}^{\mathbb{N}}: f(x) = z\}$ being dense implies that it has an element starting with $z$. $\blacksquare$

From the proof it can be seen that if we weaken the uniformity condition on $f$ so that it produces the same element for the sequences that are $Fin(\mathbb{N})$-equivalent, we still cannot diagonalize in a Borel way.

The most general version of this theorem is the following

> **Theorem**. Let $X$ be a standard Borel space and $E \subseteq X^2$ be a Borel equivalence relation. For any $x,y \in X^{\omega}$, define $x E^+ y$ if and only if $\{[x_n]: n \in \omega\}=\{[y_n]: n \in \omega\}$. Then, there does not exist a Borel map $f: X^{\omega} \rightarrow X$ such that $\neg f(x) E x_n$ for all $n \in \omega$ and $x E^+ y$ implies $f(x) E f(y).$

In some sense, this theorem tells you that you cannot diagonalize "against a Borel equivalence relation" in a "uniform" Borel way.

The operation $E \mapsto E^+$ is known as the Friedman-Stanley jump. For any Borel equivalence relation $E$ with more than one equivalence class on a standard Borel space, $E <_B E^+$ where $\leq_B$ denotes the Borel reducibility. Indeed, Friedman's own proof that $^+$ is a jump operator makes use of the Borel diagonalization theorem.

As a last remark, I should add that in the above paper Friedman claims that in order to prove this Borel diaganolization theorem in its full form, it is necessary to use $\omega_1$ many iterations of the power set of operation (Corollary 3.4), just like Borel determinacy theorem. Indeed, according to Lemma 3.2.5, there exists a constant $n \in \omega$ such that Borel determinacy up to Borel sets of rank $ \leq \alpha+n$ implies the Borel diagonalization theorem for Borel equivalence relations of rank $\leq \alpha$.