I am not sure if this works, but I believe that if $X$ is compact, one can prove that if $f^{-1}(y)$ is compact connected and of diameter bigger or equal to $\delta>0$ for every $y\in f(X)$ then $dim(f(X))< dim(X)$.

The proof goes by induction in the dimension of $f(X)$: If it is zero, it is clear (since every compact connected set of diameter bigger than $0$ has dimension $\geq 1$.

Assuming it holds for $k < n$ consider $f(X)$ of dimension $n$ and consider countably many disjoint subsets of dimension $n-1$, so, the preimage has dimension at least $n$. Assuming that $X$ has dimension $\leq n$ we get countably many disjoint open sets contradicting the compacity of $X$.

I would also bet that if $f^{-1}(y)$ is compact and connected por every $y\in f(X)$ then we should have $dim(f(X))\leq dim(X)$ (and I guess that compactness should not be necessary here).

I am not sure if this works, but I believe that if $X$ is compact, one can prove that if $f^{-1}(y)$ is compact connected and of diameter bigger or equal to $\delta>0$ for every $y\in f(X)$ then $dim(f(X))< dim(X)$.

The proof goes by induction in the dimension of $f(X)$: If it is zero, it is clear (since every compact connected set of diameter bigger than $0$ has dimension $\geq 1$.

Assuming it holds for $k < n$ consider $f(X)$ of dimension $n$ and consider countably many disjoint subsets of dimension $n-1$, so, the preimage has dimension at least $n$. Assuming that $X$ has dimension $\leq n$ we get countably many disjoint open sets contradicting the compacity of $X$.

EDIT: I have recently found this paper were the notion of cell-like mapping is discussed. There it is shown that this result does not hold unless more hypothesis are added. There is a nice discusion on the possibility of a cell-like mapping increasing dimension. See section 3.

I am not sure if this works, but I believe that if $X$ is compact, one can prove that if $f^{-1}(y)$ is compact connected and of diameter bigger or equal to $\delta>0$ for every $y\in f(X)$ then $dim(f(X))< dim(X)$.

The proof goes by induction in the dimension of $f(X)$: If it is zero, it is clear (since every compact connected set of diameter bigger than $0$ has dimension $\geq 1$.

Assuming it holds for $k < n$ consider $f(X)$ of dimension $n$ and consider countably many disjoint subsets of dimension $n-1$, so, the preimage has dimension at least $n$. Assuming that $X$ has dimension $\leq n$ we get countably many disjoint open sets contradicting the compacity of $X$.

I am not sure if this works, but I believe that if $X$ is compact, one can prove that if $f^{-1}(y)$ is compact connected and of diameter bigger or equal to $\delta>0$ for every $y\in f(X)$ then $dim(f(X))< dim(X)$.

The proof goes by induction in the dimension of $f(X)$: If it is zero, it is clear (since every compact connected set of diameter bigger than $0$ has dimension $\geq 1$.

Assuming it holds for $k < n$ consider $f(X)$ of dimension $n$ and consider countably many disjoint subsets of dimension $n-1$, so, the preimage has dimension at least $n$. Assuming that $X$ has dimension $\leq n$ we get countably many disjoint open sets contradicting the compacity of $X$.

I would also bet that if $f^{-1}(y)$ is compact and connected por every $y\in f(X)$ then we should have $dim(f(X))\leq dim(X)$ (and I guess that compactness should not be necessary here).