When does a pair of homotopic Lipschitz functions fail to admit a Lipschitz homotopy?

There is no "justice" in the world of arbitrary metric spaces:-) Let your first space $M$ be the ordinary circle with its ordinary metric, and the second space $N$ be homeomorphic to a cylinder with two border circles. Let the metric be usual on the border circles, but very rough (bumpy) on the middle part of the cylinder. You can certainly make it so rough that intermediate curves between the border circles will simply have infinite length... And of course you can make this example compact.

There are many situations where you can get Lipshitz homotopies, at least with restrictions on the topology of the spaces in question. See the paper "Quantitative algebraic topology and Lipschitz homotopy" by Ferry and Weinberger, available here.

They discuss a number of results and conjectures. One relevant one is the following, which is Theorem 1 in their paper. In it, simplicial complexes are given the path metric obtained by giving each simplex the standard Euclidean metric:

Theorem : Let $Y$ be a finite simplicial complex with finite homotopy groups in dimensions less than or equal to $d$. Then there exists $C(d)$ so that for all simplicial complexes $X$ of dimension less than or equal to $d$, if $f,g : X \rightarrow Y$ are homotopic $L$-Lipschitz maps with $L \geq 1$, then there is a $C(d)L$-Lipschitz homotopy $F$ from $f$ to $g$.

Conversely, if a Lipschitz homotopy always exists, then the homotopy groups of $Y$ are finite in dimensions less than or equal to $d$.

Let me give the answer in a clear and clean way.

Define $$ \forall_{\,x\ y\in\mathbb R}\ d(x\ y)\ :=\ |x-y|^\frac 12 $$

Thus $\ d(x\ x)=0\ $ and $\ d(x\ y) = d(y\ x)\ $ and

$$ (d(x\ y) + d(y\ z))^2 \ge |x-y|+|y-z| \ge |x-z| = d^2(x\ z) $$

thus $\ d(x\ y)+d(y\ z)\ge d(x\ z);\ $ function $\ d\ $ is a metric in $\mathbb R.\ $ And so is the induced metric in $\ \mathbb J:=[0;1].$

Thus let $\ M:=\{p\}\ $ be a 1-point metric space, and $\ N:=\mathbb J\ $ with metric $\ d\,|\,\mathbb J.$ Next, let the functions $\ f\ g: M\rightarrow N\ $ be given by $\ f(p):= 0\ $ and $\ g(p) := 1.\ $ These two functions are (topologically) homotopic.

Of course my spaces $\ M\ N\ $ are compact. But there is no Lipschytz homotopy as defined in the Question above.

INDEED:

let the Lipschytz constant $\ \kappa\ge 0\ $ be arbitrary. Then $\ \kappa\le n\ $ for a certain natural $\ n.\ $ Consider a homotopy $\ H:M\times[0;1]\rightarrow N\ $ between $\ f\ $ and $\ g.\ $ Thus $\ H(p,\ 0) = 0\ $ and $\ H(p,\ 1) = 1.$

Thus there exists an integer $\ k\ $ such that $\ 1\le k\le n^2\ $ and

$$ \left|H(p,\ \frac{k-1}{n^2}) - H(p,\ \frac k{n^2})\right|\ \ge\ \frac 1{n^2} $$

However

$$ d\left(H\left(p,\ \frac{k-1}{n^2}\right), \ H\left(p,\ \frac k{n^2}\right)\right)\ \ge\ \frac 1n \ \ge\ \kappa\cdot\left|\frac k{n^2}-\frac{k-1}{n^2}\right| $$

QED