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The answer to this question is negative even in dimension $n=1$.

A suitable counterexample can be constructed as follows.

By a result of Dranishnikov, there exists a self-map $f:M_1\to M_1$ of the Menger cube $M_1$ such that for any $y\in M_1$ the preimage $f^{-1}(y)$ is homeomorphic to $M_1$. Now take any topological copy $Y\subset M_1$ of $[0,1]$ and consider the preimage $X:=f^{-1}(Y)$. Fix a countable dense subset $Q\subset Y$ and repeating the construction of the Bernstein set, construct a subset $B\subset f^{-1}(Y\setminus Q)$ such that $f{\restriction}B$ is injective and $B$ intersects each compact set $K\subset X$ that has uncountable image $f(K)$. The zero-dimensionality of $Y\setminus Q$ and the injectivity of the map $f{\restriction}B:B\to Y\setminus Q$ implies that the space $B$ is totally disconnected.

We claim that $B$ cannot be embedded into the product $C\times[0,1]$ of the Cantor set and the interval. Assuming that there exists an embedding $h:B\to C\times[0,1]$, we can use the Lavrentiev Theorem to extend $e$ to a topological embedding $\bar h:G\to C\times[0,1]$ of some $G_\delta$-set $G\subset f^{-1}(Y\setminus Q)$ of $X$. The construction fo $B$ ensures that the projection $f(X\setminus G)$ is countable. So, we can find a point $y\in Y\setminus f(X\setminus G)$. For this point the preimage $f^{-1}(y)$ is contained in $G$ and hence admits an embedding into $C\times[0,1]$, which is not possible as $f^{-1}(y)$ is homeomorphic to $M_1$ (which does not embed into $[0,1]$).

Remark. The constructed example is essentially non-Borel. It would be interesting to know what happens for totally disconnected Polish spaces of dimension 1. Do they embed into the product $C\times[0,1]$?

The answer to this question is negative even in dimension $n=1$.

A suitable counterexample can be constructed as follows.

By a result of Dranishnikov, there exists a self-map $f:M_1\to M_1$ of the Menger cube $M_1$ such that for any $y\in M_1$ the preimage $f^{-1}(y)$ is homeomorphic to $M_1$. Now take any topological copy $Y\subset M_1$ of $[0,1]$ and consider the preimage $X:=f^{-1}(Y)$. Fix a countable dense subset $Q\subset Y$ and repeating the construction of the Bernstein set, construct a subset $B\subset f^{-1}(Y\setminus Q)$ such that $f{\restriction}B$ is injective and $B$ intersects each compact set $K\subset X$ that has uncountable image $f(K)$. The zero-dimensionality of $Y\setminus Q$ and the injectivity of the map $f{\restriction}B:B\to Y\setminus Q$ implies that the space $B$ is totally disconnected.

We claim that $B$ cannot be embedded into the product $C\times[0,1]$ of the Cantor set and the interval. Assuming that there exists an embedding $h:B\to C\times[0,1]$, we can use the Lavrentiev Theorem to extend $e$ to a topological embedding $\bar h:G\to C\times[0,1]$ of some $G_\delta$-set $G\subset f^{-1}(Y\setminus Q)$ of $X$. The construction fo $B$ ensures that the projection $f(X\setminus G)$ is countable. So, we can find a point $y\in Y\setminus f(X\setminus G)$. For this point the preimage $f^{-1}(y)$ is contained in $G$ and hence admits an embedding into $C\times[0,1]$, which is not possible as $f^{-1}(y)$ is homeomorphic to $M_1$ (which does not embed into $[0,1]$).

Remark 1. The constructed example is essentially non-Borel. It would be interesting to know what happens for totally disconnected Polish spaces of dimension 1. Do they embed into the product $C\times[0,1]$?

Remark 2. A topological space $X$ is called almost zero-dimensional if it has a base of the topology consiting of open sets $U$ whose closures $\bar U$ are intersections of open-and-closed sets. By a result of Dijsktra and van Mill each almost zero-dimensional space embeds into the complete Erdos space and the latter space embeds into $C\times[0,1]$.

The answer to this question is negative even in dimension $n=1$.

A suitable counterexample can be constructed as follows.

By a result of Dranishnikov, there exists a self-map $f:M_1\to M_1$ of the Menger cube $M_1$ such that for any $y\in M_1$ the preimage $f^{-1}(y)$ is homeomorphic to $M_1$. Now take any topological copy $Y\subset M_1$ of $[0,1]$ and consider the preimage $X:=f^{-1}(Y)$. Fix a countable dense subset $Q\subset Y$ and repeating the construction of the Bernstein set, construct a subset $B\subset f^{-1}(Y\setminus Q)$ such that $f{\restriction}B$ is injective and $B$ intersects each compact set $K\subset X$ that has uncountable image $f(K)$. The zero-dimensionality of $Y\setminus Q$ and the injectivity of the map $f{\restriction}B:B\to Y\setminus Q$ implies that the space $B$ is totally disconnected.

We claim that $B$ cannot be embedded into the product $C\times[0,1]$ of the Cantor set and the interval. Assuming that there exists an embedding $h:B\to C\times[0,1]$, we can use the Lavrentiev Theorem to extend $e$ to a topological embedding $\bar h:G\to C\times[0,1]$ of some $G_\delta$-set $G\subset f^{-1}(Y\setminus Q)$ of $X$. The construction fo $B$ ensures that the projection $f(X\setminus G)$ is countable. So, we can find a point $y\in Y\setminus f(X\setminus G)$. For this point the preimage $f^{-1}(y)$ is contained in $G$ and hence admits an embedding into $C\times[0,1]$, which is not possible as $f^{-1}(y)$ is homeomorphic to $M_1$ (which does not embed into $[0,1]$).

The answer to this question is negative even in dimension $n=1$.

A suitable counterexample can be constructed as follows.

By a result of Dranishnikov, there exists a self-map $f:M_1\to M_1$ of the Menger cube $M_1$ such that for any $y\in M_1$ the preimage $f^{-1}(y)$ is homeomorphic to $M_1$. Now take any topological copy $Y\subset M_1$ of $[0,1]$ and consider the preimage $X:=f^{-1}(Y)$. Fix a countable dense subset $Q\subset Y$ and repeating the construction of the Bernstein set, construct a subset $B\subset f^{-1}(Y\setminus Q)$ such that $f{\restriction}B$ is injective and $B$ intersects each compact set $K\subset X$ that has uncountable image $f(K)$. The zero-dimensionality of $Y\setminus Q$ and the injectivity of the map $f{\restriction}B:B\to Y\setminus Q$ implies that the space $B$ is totally disconnected.

We claim that $B$ cannot be embedded into the product $C\times[0,1]$ of the Cantor set and the interval. Assuming that there exists an embedding $h:B\to C\times[0,1]$, we can use the Lavrentiev Theorem to extend $e$ to a topological embedding $\bar h:G\to C\times[0,1]$ of some $G_\delta$-set $G\subset f^{-1}(Y\setminus Q)$ of $X$. The construction fo $B$ ensures that the projection $f(X\setminus G)$ is countable. So, we can find a point $y\in Y\setminus f(X\setminus G)$. For this point the preimage $f^{-1}(y)$ is contained in $G$ and hence admits an embedding into $C\times[0,1]$, which is not possible as $f^{-1}(y)$ is homeomorphic to $M_1$ (which does not embed into $[0,1]$).

Remark. The constructed example is essentially non-Borel. It would be interesting to know what happens for totally disconnected Polish spaces of dimension 1. Do they embed into the product $C\times[0,1]$?

The answer to this question is negative even in dimension $n=1$.

A suitable counterexample can be constructed as follows.

By a result of Dranishnikov, there exists a self-map $f:M_1\to M_1$ of the Menger cube $M_1$ such that for any $y\in M_1$ the preimage $f^{-1}(y)$ is homeomorphic to $M_1$. Now take any topological copy $Y\subset M_1$ of $[0,1]$ and consider the preimage $X:=f^{-1}(Y)$. Fix a countable dense subset $Q\subset Y$ and repeating the construction of the Bernstein set, construct a subset $B\subset f^{-1}(Y\setminus Q)$ such that $f{\restriction}B$ is injective and $B$ intersects each compact set $K\subset X$ that has uncountable image $f(K)$. The zero-dimensionality of $Y\setminus Q$ and the injectivity of the map $f{\restriction}B:B\to Y\setminus Q$ implies that the space $B$ is totally disconennected.

We claim that $B$ cannot be embedded into the product $C\times[0,1]$ of the Cantor set and the interval. Assuming that there exists an embedding $h:B\to C\times[0,1]$, we can use the Lavrentiev Theorem to extend $e$ to a topological embedding $\bar h:G\to C\times[0,1]$ of some $G_\delta$-set $G\subset f^{-1}(Y\setminus Q)$ of $X$. The construction fo $B$ ensures that the projection $f(X\setminus G)$ is countable. So, we can find a point $y\in Y\setminus f(X\setminus G)$. For this point the preimage $f^{-1}(y)$ is contained in $G$ and hence admits an embedding into $C\times[0,1]$, which is not possible as $f^{-1}(y)$ is homeomorphic to $M_1$ (which does not embed into $M_1$).

The answer to this question is negative even in dimension $n=1$.

A suitable counterexample can be constructed as follows.

By a result of Dranishnikov, there exists a self-map $f:M_1\to M_1$ of the Menger cube $M_1$ such that for any $y\in M_1$ the preimage $f^{-1}(y)$ is homeomorphic to $M_1$. Now take any topological copy $Y\subset M_1$ of $[0,1]$ and consider the preimage $X:=f^{-1}(Y)$. Fix a countable dense subset $Q\subset Y$ and repeating the construction of the Bernstein set, construct a subset $B\subset f^{-1}(Y\setminus Q)$ such that $f{\restriction}B$ is injective and $B$ intersects each compact set $K\subset X$ that has uncountable image $f(K)$. The zero-dimensionality of $Y\setminus Q$ and the injectivity of the map $f{\restriction}B:B\to Y\setminus Q$ implies that the space $B$ is totally disconnected.

We claim that $B$ cannot be embedded into the product $C\times[0,1]$ of the Cantor set and the interval. Assuming that there exists an embedding $h:B\to C\times[0,1]$, we can use the Lavrentiev Theorem to extend $e$ to a topological embedding $\bar h:G\to C\times[0,1]$ of some $G_\delta$-set $G\subset f^{-1}(Y\setminus Q)$ of $X$. The construction fo $B$ ensures that the projection $f(X\setminus G)$ is countable. So, we can find a point $y\in Y\setminus f(X\setminus G)$. For this point the preimage $f^{-1}(y)$ is contained in $G$ and hence admits an embedding into $C\times[0,1]$, which is not possible as $f^{-1}(y)$ is homeomorphic to $M_1$ (which does not embed into $[0,1]$).