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Just a small addition to the existing answers.

Theorem. There exists a non-metrizable compact Hausdorff space $K$ admitting a continuous surjective function $f:K\to[0,1]^2$ to the unit square such that $f$ has a Borel section if and only if CH holds. Then taking the Peano square-filling curve $p:[0,1]\to[0,1]^2$ we can show that $p$ has a Borel $f$-lifting $[0,1]\to K$ if and only if the Continuum Hypothesis holds.

Skecth of the proof. Consider the split interval $\ddot{\mathbb I}$ (which is the lexicographic product $[0,1]\times\{0,1\}$ endowed with the order topology). It is well-known that $\ddot{\mathbb I}$ is a compact Hausdorff space admitting a continuous surjective map $\ddot{\mathbb I}\to [0,1]$. Then the square $K=\ddot{\mathbb I}\times\ddot{\mathbb I}$ admits a continuous surjective map $f:K\to[0,1]^2$ such that for every point $z\in[0,1]^2$ the preimage $f^{-1}(z)$ contains at most 4 points. Using the answer (and comments) to this question, it can be shown that $f$ has the desired property: it has a Borel section if and only if CH holds. $\quad\square$

The above theorem suggests the following (open?)

Problem. Is there a 2-to-1 map $f:K\to M$ from a compact (Rosenthal) space onto a compact metrizable space $M$, which has no Borel selections?

Just a small addition to the existing answers.

Theorem. There exists a non-metrizable compact Hausdorff space $K$ admitting a continuous surjective function $f:K\to[0,1]^2$ to the unit square such that $f$ has a Borel section if and only if CH holds. Then taking the Peano square-filling curve $p:[0,1]\to[0,1]^2$ we can show that $p$ has a Borel $f$-lifting $[0,1]\to K$ if and only if the Continuum Hypothesis holds.

Skecth of the proof. Consider the split interval $\ddot{\mathbb I}$ (which is the lexicographic product $[0,1]\times\{0,1\}$ endowed with the order topology). It is well-known that $\ddot{\mathbb I}$ is a compact Hausdorff space admitting a continuous surjective map $\ddot{\mathbb I}\to [0,1]$. Then the square $K=\ddot{\mathbb I}\times\ddot{\mathbb I}$ admits a continuous surjective map $f:K\to[0,1]^2$ such that for every point $z\in[0,1]^2$ the preimage $f^{-1}(z)$ contains at most 4 points. Using the answer (and comments) to this question, it can be shown that $f$ has the desired property: it has a Borel section if and only if CH holds. More details on the proof can be found in this paper. $\quad\square$

The above theorem suggests the following (open?)

Problem. Is there a 2-to-1 map $f:K\to M$ from a compact (Rosenthal) space onto a compact metrizable space $M$, which has no Borel selections?

Just a small addition to the existing answers.

Theorem. There exists a non-metrizable compact Hausdorff space $K$ admitting a continuous surjective function $f:K\to[0,1]^2$ to the unit square such that $f$ has a Borel section if and only if CH holds. Then taking the Peano square-filling curve $p:[0,1]\to[0,1]^2$ we can show that $p$ has a Borel $f$-lifting $[0,1]\to K$ if and only if the Continuum Hypothesis holds.

Skecth of the proof. Consider the Alexandrov two-arrow space $A$ (which is the lexicographic product $[0,1]\times\{0,1\}$ endowed with the order topology). It is well-known that $A$ is a compact Hausdorff space admitting a continuous surjective map $A\to [0,1]$. Then the square $K=A\times A$ admits a continuous surjective map $f:K\to[0,1]^2$ such that for every point $z\in[0,1]^2$ the preimage $f^{-1}(z)$ contains at most 4 points. Using the answer (and comments) to this question, it can be shown that $f$ has the desired property: it has a Borel section if and only if CH holds. $\quad\square$

The above theorem suggests the following (open?)

Problem. Is there a 3-to-1 map $f:K\to M$ from a compact (Rosenthal) space onto a compact metrizable space $M$, which has no Borel selections?

More details (with proofs of the above theorems) could be written in a separate paper, if there is such a desire or necessity.

Just a small addition to the existing answers.

Theorem. There exists a non-metrizable compact Hausdorff space $K$ admitting a continuous surjective function $f:K\to[0,1]^2$ to the unit square such that $f$ has a Borel section if and only if CH holds. Then taking the Peano square-filling curve $p:[0,1]\to[0,1]^2$ we can show that $p$ has a Borel $f$-lifting $[0,1]\to K$ if and only if the Continuum Hypothesis holds.

Skecth of the proof. Consider the split interval $\ddot{\mathbb I}$ (which is the lexicographic product $[0,1]\times\{0,1\}$ endowed with the order topology). It is well-known that $\ddot{\mathbb I}$ is a compact Hausdorff space admitting a continuous surjective map $\ddot{\mathbb I}\to [0,1]$. Then the square $K=\ddot{\mathbb I}\times\ddot{\mathbb I}$ admits a continuous surjective map $f:K\to[0,1]^2$ such that for every point $z\in[0,1]^2$ the preimage $f^{-1}(z)$ contains at most 4 points. Using the answer (and comments) to this question, it can be shown that $f$ has the desired property: it has a Borel section if and only if CH holds. $\quad\square$

The above theorem suggests the following (open?)

Problem. Is there a 2-to-1 map $f:K\to M$ from a compact (Rosenthal) space onto a compact metrizable space $M$, which has no Borel selections?

Just a small addition to the existing answers.

Theorem. There exists a non-metrizable compact Hausdorff space $K$ admitting a continuous surjective function $f:K\to[0,1]^2$ to the unit square such that $f$ has a Borel section if and only if CH holds. Then taking the Peano square-filling curve $p:[0,1]\to[0,1]^2$ we can show that $p$ has a Borel $f$-lifting $[0,1]\to K$ if and only if the Continuum Hypothesis holds.

Skecth of the proof. Consider the Alexandrov two-arrow space $A$ (which is the lexicographic product $[0,1]\times\{0,1\}$ endowed with the order topology). It is well-known that $A$ is a compact Hausdorff space admitting a continuous surjective map $A\to [0,1]$. Then the square $K=A\times A$ admits a continuous surjective map $f:K\to[0,1]^2$ such that for every point $z\in[0,1]^2$ the preimage $f^{-1}(z)$ contains at most 4 points. Using the answer (and comments) to this question, it can be shown that $f$ has the desired property: it has a Borel section if and only if CH holds. $\quad\square$

This example can be modified to construct a 3-to-1 map $p:K\to D$ of a Rosenthal compact $K$ onto the unit disk $D=\{z\in\mathbb C:|z|\le1\}$ that has a Borel selection if and only if CH holds.

To construct the compact $K$, consider the three closed sectors $S_0,S_1,S_2\subset \mathbb C$, where $$S_k=\{z\in\mathbb C:\tfrac23\pi k\le\arg(z)\le\tfrac23\pi(k+1)\}\mbox{ for }k\in\{0,1,2\}.$$ For every point $z\in D$ and every $k\in\{0,1,2\}$ let $S_k(z)=D\cap(z+S_k)$. For a subset $A\subset D$ let $\chi_A:D\to\{0,1\}$ be the characteristic function of the set $A$ in $A$, i.e., a unique function such that $\chi_A^{-1}(1)=1$.

It can be shown that the subset $$K=\{\chi_{S_k(z)}:k\in\{0,1,2\},\;z\in D\}\subset \{0,1\}^D$$is Rosenthal compact and the map $$p:K\to D,\;\;p:\chi_{S_k(z)}\mapsto z,$$ is continuous and 3-to-1 (which means that $|p^{-1}(z)|=3$ for any $z\in D$).

We say that a function $s:D\to K$ is a selection of the map $p$ if $p\circ s$ is the identity map of $D$.

By analogy with the above theorem, one can prove the following

Theorem 3. For the 3-to-1 map $p:K\to D$ the following conditions are equivalent:

(1) $p$ has a Borel selection $s:D\to K$;

(2) $p$ has an $F_\sigma$-measurable selection;

(3) CH holds.

The above theorems suggest the following (open?)

Problem. Is there a 2-to-1 map $f:K\to M$ from a compact (Rosenthal) space onto a compact metrizable space $M$, which has no Borel selections?

More details (with proofs of the above theorems) could be written in a separate paper, if there is such a desire or necessity.

Just a small addition to the existing answers.

Theorem. There exists a non-metrizable compact Hausdorff space $K$ admitting a continuous surjective function $f:K\to[0,1]^2$ to the unit square such that $f$ has a Borel section if and only if CH holds. Then taking the Peano square-filling curve $p:[0,1]\to[0,1]^2$ we can show that $p$ has a Borel $f$-lifting $[0,1]\to K$ if and only if the Continuum Hypothesis holds.

Skecth of the proof. Consider the Alexandrov two-arrow space $A$ (which is the lexicographic product $[0,1]\times\{0,1\}$ endowed with the order topology). It is well-known that $A$ is a compact Hausdorff space admitting a continuous surjective map $A\to [0,1]$. Then the square $K=A\times A$ admits a continuous surjective map $f:K\to[0,1]^2$ such that for every point $z\in[0,1]^2$ the preimage $f^{-1}(z)$ contains at most 4 points. Using the answer (and comments) to this question, it can be shown that $f$ has the desired property: it has a Borel section if and only if CH holds. $\quad\square$

The above theorem suggests the following (open?)

Problem. Is there a 3-to-1 map $f:K\to M$ from a compact (Rosenthal) space onto a compact metrizable space $M$, which has no Borel selections?

More details (with proofs of the above theorems) could be written in a separate paper, if there is such a desire or necessity.