= Solution
Take $M=\mathbb R$ and the constant <contraction mapping> $f(x)=0$. For $p=(0,0)$, $q=(0,1)$,
$$
\delta(p,q)=1,\qquad F(p)=(0,0),\quad F(q)=(1,0),\qquad\delta(F(p),F(q))=1.
$$
No Lipschitz constant less than one can work, so \b[$F$ need not be a contraction]. More generally, differing only in the second input coordinate already prevents a strict contraction whenever $M$ has two points.
If $f$ has contraction constant $0\leq c<1$, two iterations give
$$
F^2(m,m')=(f(m),f(m')),
$$
and consequently
$$
\boxed{\delta(F^2(p),F^2(q))\leq c\,\delta(p,q).}
$$
Thus \b[$F^2$ is a <contraction mapping>]. The <swap lift of a contraction> shows why an iterate can contract even when one step does not.
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