Let $n$ be any natural number, $X$ be the space $\mathbb R^n$, $M$ be a nonempty compact subset of the space $X$, and $p_M$ be the metric projection associated with $M$. As I understood the definition of upper hemicontinuity
from Wikipedia, the function $p_M$ is upper hemicontinuous, that is for each $a\in X$ and each open set $V\supset p_M(a)$, there exists a neighbourhood $U$ of $a$ such that $p_M(x)\subset V$ for all $x\in U$.
Indeed, otherwise there exists a sequence $(a_n)_{n\in\mathbb N}$ of points of $X$ convergent to $a$ and
a sequence $(b_n)_{n\in\mathbb N}$ such that $b_n\in p_M(a_n)\setminus V$ for each natural $n$. Since the set $M$ is a metric compact, taking a subsequence of the sequence $(a_n)_{n\in\mathbb N}$, if needed, we can suppose that the sequence $(b_n)_{n\in\mathbb N}$ converges to a point $b\in M$. Let $D=d(a,M)$, $D'=d(a,b)$, and for each natural $n$ let $D_n=d(a_n,M)$. The triangle inequality implies that
$|D-D_n|=|d(a,M)-d(a_n,M)|\le d(a,a_n)$. Since the sequence $(a_n)_{n\in\mathbb N}$ converges to $a$,
the sequence $(D_n)_{n\in\mathbb N}$ converges to $D$. For each natural $n$ we have $b_n\in p_M(a_n)$, so
$d(a_n,b_n)=d(a_n,M)=D_n$. Since the sequence $(a_n)_{n\in\mathbb N}$ converges to $a$ and
the sequence $(b_n)_{n\in\mathbb N}$ converges to $b$, the sequence
$(D_n)_{n\in\mathbb N}=(d(a_n,b_n))_{n\in\mathbb N}$ converges to $d(a,b)=D'$. Thus $D'=D$ and thus $d(a,M)=d(a,b)$, so $d\in p_M(a)$, a contradiction.
For each $a\in X$ the set $p_M(a)$ is a convex compact as the intersection of the convex compact sets $M$ and the $\ell_1$-ball of radius $d(a,M)$ centered at $a$. Then according to Wikipedia, the set $p_M(a)$ is Chebyshev with respect to the Euclidean metric $d_E$ on $\mathbb R^n$, that is there exists a unique point $b(a)\in p_M(a)$ such that $d_E(a,b(a))=d_E(a,M)$.
Unfortunately, as I understood the definition of lower hemicontinuity
from Wikipedia, the function $p_M$ can fail to be lower hemicontinuous. Namely, let $n=3$, $M$ be a cone, whose vertex is $(0,0,1)$ and base is the unit circle in the $xOy$ plane centered at $(2,0,0)$. Then $p_M(0,0,0)$ is the segment with the endpoints $(0,0,1)$ and $(1,0,0)$. Let $y>0$ and $P=(0,y,0)$.
Let us study the set $p_M(P)$.
The distance $d(P,M)$ from the point $P$ to $M$ equals the distance from $P$ to the lateral surface of the cone $M$, whose points $Q$ can be parametrized as $(1-h)(0,0,1)+h(2+cos t,\sin t,0)$, $h\in [0,1]$, $t\in [0,2\pi]$. Then $d(P,Q)=1+h(1+\cos t)+|y-h\sin t|$. If $h=0$ then $d(P,Q)=1+y$. Otherwise the minimum of $d(P,Q)$ can be attained either when the partial derivative $d(P,Q)_t$ of $d(P,Q)$ with respect to $t$ either equals zero of does not exist.
In the first case we have $\cos t=\pm \sin t=\pm \frac{\sqrt{2}}2$. To minimize $d(P,Q)$ we need $-\cos t=\sin t=\frac{\sqrt{2}}2$. If $h\le \sqrt{2}y$ then
$$d(P,Q)=1+h\left(1-\frac{\sqrt{2}}2\right)+y-h\cdot \frac{\sqrt{2}}2=1+h(1-\sqrt{2})+y$$
and its minimum $1+(\sqrt{2}-1)y$ is attained when $h=\sqrt{2}y$.
If $h\le \sqrt{2}y$ then
$$d(P,Q)=1+h\left(1-\frac{\sqrt{2}}2\right)+h\cdot \frac{\sqrt{2}}2-y=1+h+y$$
and its minimum $1+y$ is attained when $h=0$.
In the second case we have $y=h\sin t<h$, so $\cos t=\pm \sqrt{1-\left(\frac yh\right)^2}$. To minimize $d(P,Q)$ we need to choose the minus sign here. Then
$$d(P,Q)=1+h\left(1-\sqrt{1-\left(\frac yh\right)^2}\right)=1+h-\sqrt{h^2-y^2}.$$
The partial derivative $d(P,Q)_t$ of $d(P,Q)$ with respect to $t$ equals
$$1-\frac 1{\sqrt{1-\left(\frac yh\right)^2}}<0,$$
so to attain the minimum of $d(P,Q)$ we need to put $h=1$. Then $d(P,Q)=2-\sqrt{1-y^2}$.
From above we see that for the point $Q$, $h\in\{0,1\}$.
But even when the function $p_M$ is not lower hemicontinous, there still is a hope that it has a single-valued continuous selection. I am going to think about this more.
