Is there a compact metric space $X$ of covering dimension $2$ without a Lipschitz surjection on $[0,1]^2$?
For a space $X$ with Hausdorff dimension greater than $2$, we have a negative answer (see Theorem 1.1 in article https://arxiv.org/pdf/1203.0686.pdf).
Let $\ n\in\mathbb N.\ $ It's convenient to consider the injective metric in $\ \mathbb R^n,\ $ it's Lipschitz equivalent to the Euclidean metrics.
By dimension, let's mean the topological dimension dim (say, covering -- for metric compact spaces, topological has only one standard meaning).
Theorem For every metric compact space $\ X,\ $ with a topologically agreeing metrics $\ d,\ $ such that $\ \dim(X)\ge n,\ $ there exists a Lipschitz surjection $\ f:X\to[-1;1]^n.$
Proof Due to the dimension $\ \ge n\ $ there are $\ (F_k:\ k=\pm1 \ldots \pm n)\,\ $ closed subsets of $\ X\ $ such that
$$ \forall_{k=1}^n\quad F_{-k}\cap F_k\ =\ \emptyset $$ and $$ \bigcap_{k=1}^n\,S_k\ \ne\ \emptyset $$ for arbitrary closed separators S_k between $\ F_{-k}\ $ and $\ F_k,\ $ where $\ 1\le k\le n.$
Let $\ g_k:F_{-k}\cup F_k\to[-1;1]\ $ be defined by $\ g_k(x)=-1\ $ for every $\ x\in F_{-k},\ $ and $\ g_k(x)=1\ $ for every $\ x\in F_k,\ $ whenever $\ 1\le k\le n.\ $ These maps are obviously Lipschitz. Since $\ [-1;1]\ $ is metrically injective (is a metric AR), there are Lipschitz maps $\ f_k:X\to [-1;1]\ $ that extend the respective $\ g_k,\ $, hence
$$ f\ :=\ \triangle_{k=1}^n f_k:X\to[-1;1]^n $$ is a Lipschitz map.
Such continuous maps that start with the non-separable collections $\ F_{\pm k}\ $ are well-known and easily seen to be surjective (or even universal).
End of Proof
Here is a sketched alternative perspective on why this is impossible, i.e., why every compact metric space of topological dimension $n$ has a Lipschitz map onto $[0,1]^n$.
If $X$ is compact with topological dimension $n$, then it admits a continuous map $f$ to $[0,1]^n$ with a stable value. This means a value $y\in\mathbb{R}^n$ such that every continuous map from $X$ to $[0,1]^n$ that is sufficiently close to $f$ in the supremum distance must contain $y$ in its image. (See Modern Dimension Theory by Nagata.)
In particular, this means that $f(X)$ contains an open neighborhood of $y$ in $[0,1]^n$ (otherwise you could "nudge" the map and lose $y$ from the image). The map $f$ can be approximated by Lipschitz maps into $[0,1]^n$. Choose one of these Lipschitz maps close enough to $f$; then it also will have $y$ as a stable value and thus contain a neighborhood of $y$ in its image. Once you have that, you can post-compose with a Lipschitz map that sends a small ball around $y$ onto $[0,1]^n$ and the rest of the cube to the boundary.
Probably Wlod AA's answer cuts more to the heart of why there is a map with a stable value to begin with, but I am less familiar with this.
