Let $P$ be the point in/on the polyhedron closest to $K$. Observe that $P$ must lie on the polyhedron, otherwise we could intersect $KP$ with the polyhedron. We show that $P$ is the unique point which we want. The only point belonging to the ball with diametre $PK$ and the polyhedron is $P$. We now need to show that $P$ belongs to all the other balls. To do this, construct a plane $J$ which passes through $P$ and is perpendicular to $KP$. We first claim that the no point $S$ on the polyhedron lies on the same side of $J$ as $K$. FTSOC assume $S$ exists, and let $Y$ be the foot of the perpendicular from $K$ to $PS$. If $Y = P$, then $S$ lies on $J$, which is a contradiction. If $Y$ lies on segment $PS$, we have $KY < KP$ and $Y$ inside polyhedron since it is convex. Contradiction to definition of $P$ If $Y$ lies on ray $PS$ beyond $S$, than $KS < KP$. Contradiction to definition of $P$ If $Y$ lies on the other side of the $J$ as $K$, but we then have that $KY > \text{dist}(K,J) = KP$. Not possible for foot of perpendicular. Thus, our claim is true. Now, observe that for any point $M$ on the polyhedron, $\angle MPK \geq 90$, so $P$ belongs to the ball with diameter $MK$.

Let the polyhedron be a unit cube. $K $ is at distance $10. $ Every point in the cube is within all such balls because their radius $\approx 5\text {unit} $ A point does not necessarily exist that lies within all the balls. Let a unit cube about $0$ and place $K $ at $(0.501,0,0) $ The ball about $(-0.5,0,0) $ reaches as far $x=0.0005.$. Rule out all the points with $x >0.0005$ The ball abut $(0.5,0,0) $ is only radius $0.0005$ unit, rule out all the points with $x < 0.496$ Substitute radius for diameter in the problem.