No, the contractive property of the nearest-point projection is a special property of Hilbert spaces. Here is a
counterexample in the $3$-dimensional real space with norm $\|x\|_4=(x_1^4+x_2^4+x_3^4)^{1/4}$. This is a nice norm:
uniformly convex and uniformly smooth. The nearest-point projection onto any convex closed set is uniquely defined.
Consider the nearest-point projection $P$ onto the line $L=\{(t,t,t): t\in \mathbb R\}$. A point $x$ projects into $(0,0,0)$
if and only if the $t$-derivative of $(x_1-t)^4+(x_2-t)^4+(x_3-t)^4$ vanishes at $t=0$. Therefore, the preimage
of $(0,0,0)$ under $P$ is the surface $S_0=\{x:x_1^3+x_2^3+x_3^3=0\}$. Since the distance is translation-invariant, $P$ commutes with translations along $L$. It follows that the preimage of $(1,1,1)$ is the surface
$S_1=\{x:(x_1-1)^3+(x_2-1)^3+(x_3-1)^3=0\}$.
It remains to pick a pair of points on these two surfaces such that the distance between them is less than $\|(1,1,1)\|_4=3^{1/4}$.
For example, I took $a=(2^{1/3}+1,0,0)\in S_1$ and considered the distance from $a$ to the points
$(2^{1/3}s,-s,-s)\in S_0$. The minimum is attained around $s\approx 0.93$, and it is less than $2.9^{1/4}$.
Answering a follow-up question, I add a proof of the contractive property in Hilbert spaces.
Let $P$ be projection onto a closed convex set $C$, and consider $a'=P(a)$ and $b'=P(b)$.
Since the segment $a'b'$ is contained in $C$, we have $\|(1-t)a'+tb'-a\|\ge \|a'-a\|$ for $t\in [0,1]$.
Therefore,
$$0\le \frac{d}{dt}\|(1-t)a'+tb'-a\|^2\bigg|_{t=0} = 2 \langle b'-a' , a'-a \rangle \tag{1}$$
Similarly,
$$\langle a'-b' , b'-b \rangle \ge 0\tag{2}$$
Now consider the function
$$
d(t)=\|(1-t)a'+ta - ((1-t)b'+tb)\|^2 = \|(a'-b') + t (a-a'-b+b') \|^2
$$
which is a quadratic polynomial with nonnegative coefficient of $t^2$. From (1) and (2) we have
$$
d\,'(0) = 2 \langle a'-b' , a-a'-b+b' \rangle \ge 0
$$
Therefore $d$ is increasing on $[0,\infty)$. In particular, $d(1)\ge d(0)$ which means
$\|a-b\|\ge \|a'-b'\|$.
Note that $\Pi_C(x^\ast) \in C\subseteq H$, and $\Pi_H(x^\ast) \in H$. Then by construction of $H$,
$$0 \geq \langle x^\ast - \Pi_C(x^\ast),\Pi_H(x^\ast) - \Pi_C(x^\ast)\rangle = \langle \Pi_C(x^\ast) - x^\ast, \Pi_C(x^\ast) - \Pi_H(x^\ast) \rangle,$$
and by the variational inequality of Euclidean distance
$$0 \geq \langle x^\ast - \Pi_H(x^\ast),\Pi_C(x^\ast) - \Pi_H(x^\ast)\rangle.$$
So add the two inequalities together, and you should see that, in fact, $\Pi_C(x^\ast) = \Pi_H(x^\ast)$!
Best Answer
The case $d_C(x^\ast)\geq\max_{C\subseteq H}d_H(x^\ast)$ is trivial.
Note that it has been shown in the preceding question that $C\subseteq H_\ast \equiv \overline{H}^-(a,\gamma)$, where $H_\ast$ is a closed half space, $a = x^\ast - \Pi_C(x^\ast)$ and $\gamma = \langle x^\ast, \Pi_C(x^\ast) \rangle$, and in fact, $d_C(x^\ast) = d_{H_\ast}(x^\ast)$. Suppose $d_C(x^\ast)\not\leq\max_{C\subseteq H}d_H(x^\ast)$. Then $d_C(x^\ast)>\max_{C\subseteq H}d_H(x^\ast)$, and it's clear we have a contradiction.