I came up with a solution which was incorrect. However, one of my partial results might be interesting (or might be trivial)Yes.
First let's sayLet me use the start ofreduction to labelings suggested by Rohit. Let's call a $2$two-bushy tree where $T$$\sigma$ is the noderoot a $\sigma$$2$-branchy tree.
Result: LetFor any continuous function $v_1$ and$L$ from $v_2$ be two nodes. There$4^{<\omega}$ to $\{0,1\}$, there is a $2$-bushy tree $T_1$ starting at $v_1$ andeither a $2$three-bushybranchy tree $T_2$ starting at $v_2$ such that either$T$ where $f$ is constant$L$ restricted to $T_1$, or $f$$[T]$ is the constant restricted tofunction $T_2$,$0$ or the image ofa two-branchy tree $f$$T$ where $L$ restricted to $T_1$$[T]$ is disjoint from the image of $f$ restricted toconstant function $T_2$$1$.
I thought that This follows by applyingbackwards induction on the last legnodes, starting from nodes where $L$ is constant. If a node has all four children the base of one of these two types of trees, then either three have the trichotomy many times you could produce an injective functionfirst type of tree or two have the second, but this turned out not to beso it has one of the casetwo types of threes.
To prove this, note:
Lemma: Let $T$ beAs a corollary there is either a $a$$2$-bushybranchy tree starting at a node $v$. Let$T$ such that $Z$ be a closed set of$L$ restricted to $2^{ <\omega}$. Suppose that$T$ is $0$, for everyor a $b$$2$-bushy subtreebranchy tree $T'$ of$T$ such that $L$ restricted to $T$ starting atis $v$$1$. Without loss of generality, assume it is $f([T']) \cap Z \neq \varnothing$$0$. Then let $k$ be the maximum number such that there is ana $a-b+1$$2$-bushy subtree $T^*$ ofbranchy tree $T$ suchwhere $f$ restricted to $[T]$ lands on functions that begin with $f(T^*) \subseteq Z$$k$ zeros. By assumption $k\geq 1$.
Proof of lemma: LetIf $T^*$ be the subtree of$k=\infty$ we win $T$ consisting of all vertices that are not the start of an- there is a tree where $b$$f$ is constant.
So assume $k$ is finite. Suppose there exists a $2$-bushy subtree whose branches are all sent tobranchy tree $2^{ \omega}-Z$ by$T'$ such that $f$ restricted to $[T']$ lands on functions that do not begin with $k$ zeros. Then every node ofwe win. Indeed the first $T^*$ has at least$k$ values of $a+b-1$ successors in$f(x_0x_1\dots)$ are determined by $T^*$$x_0x_1\dots x_{N-1}$ for some $N$. Then choose a two-branchy tree of sequences $x_0x_1\dots$ such that if it had any fewer, it would have $b$ sucessors, each$x_i$ is the startleftmost of athe two options at $b$-bushy subtree whose branches are sent to$x_0\dots x_{i-1}$ then $2^{\omega}-Z$$x_{Ni}x_{N{i+1}} \dots x_{Ni+N-1}$ is in $T$, hence it would beand if $x_i$ is the startrightmost of athe two options then $b$-bushy subtree whose branches are sent to$x_{Ni}x_{N{i+1}} \dots x_{Ni+N-1}$ is in $2^{\omega}-Z$. By assumption,$T'$ $T^*$ contains(and we might have to switch left and right for $v$$x_0$). Finally $[T^*] \subseteq Z$ because any branch not in $Z$ would have an open neighborhood of branches not inThen $Z$, in which it would be easy$f$ restricted to form a $b$-bushythat tree startingis injective, as we can reconstruct $x_i$ by looking at one of the vertices$Ni$ through $Ni+k-1$ digits of the branch whose branches do not intersect $Z$$f(x_0x_i\dots)$.
Now using the lemma, takeIf there does not exist such a subtree $T_1$ starting at $v_1$ and let $Z=f([T_1])$. Then either we can find a 2$2$-bushybranchy tree $T_2$ with $f([T_2])$ disjoint from $f([T_1])$, and we're done, or we can findthen there exists a $3$-bushybranchy tree $S_2$ with $f([S_2]) \subseteq Z$. Now pick a point in$T^*$ where $Z$. We can either find a$f$ restricted to $2$-bushy subtree$T^*$ always begins with $T_2$$k$ zeroes. Then consider the tree of $S_2$, withsequences $f([T_2])$ not containing$x$ that point, or find a 2-bushy subtree withare in $f([T_2])$ constant$T^*$ and equal to that point. In the second case, we are done. Inafter removing the first casedigit, remain in $f([T_2])$$T^*$. This is a closed subset of $f([T_1])$. Iterating, we get a 2$2$-bushy subtree $T_3$ starting at $v_1$ with $f([T_3])$ a proper closed subset of $f([T_1])$. We can do this infinitely many times and take the limitbranchy tree, and $f$ applied to the limit is containedeverything in the intersection, and then repeat the processit has $k+1$ zeroes. By transfinite inductionThis contradicts our assumption on / Zorn's Lemma, we eventually must reach a closed set containing no points, which is the empty set, a contradiction$k$.
So we're donewe win.