(For a finite set of points only.) Let $D$ be a disk of smallest radius that covers all marked points. Consider the marked points on the boundary $C$ of this disk. Note that if all marked points on $C$ lie on an arc smaller than the half circle (ASTTHC for short), then the disk can be moved a little towards these points on the boundary and its radius can be decreased. Since we assumed that our disk has minimal radius, the marked points on its boundary do not lie on an ASTTHC.

If the two endpoints of a diagonal of $D$ are marked, then $D$ is the smallest disk containing these two points, hence must have radius at most 1.

If there are 3 marked points on $C$ that do not lie on an ASTTHC, then $D$ is the smallest disk covering these 3 points and hence must have radius at most 1. (In this case the triangle formed by the three points is acute and $C$ is its circumcircle.)

If there are more than 3 marked points on the boundary that do not lie on an ASTTHC, then we can remove one of them so that the remaining points again do not lie on an ASTTHC. By induction this leads us to the case of 3 points. Indeed, given 4 or more points on $C$, choose 3 points that lie on a half circle. Then the middle point can be removed.

1. Define Notations and Assumptions:
- Let $(XYZ)$ denote the circumcircle of $\triangle XYZ$.
- Let $(XY)$ denote the circle with diameter $XY$.
- Let $r(\mathcal{C})$ denote the radius of a disc/circle $\mathcal{C}$.
- Let $d(XYZ)$ denote the radius of the smallest disc containing points $X, Y, Z$.

We may assume without loss of generality (WLOG) that the points are in a convex position. If a point lies inside some triangle, we can ignore that point.

2. Lemma:
$$d(ABC) = \begin{cases} r(ABC) & \text{if } \triangle ABC \text{ is non-obtuse} \\ \max\{r(BC), r(CA), r(AB)\} & \text{if } \triangle ABC \text{ is obtuse} \end{cases}$$
Proof: Omitted. $\square$

3. Choose Points and Define Disc:
- Pick three points $A, B, C$ such that $d(ABC)$ is the maximum possible.
- Let $\mathcal{D}$ denote the disc covering $A, B, C$.
- We will show that $\mathcal{D}$ covers all other points.

4. **Assume a Point Outside $\mathcal{D}$:**
- Assume, for contradiction, that some point $X$ lies outside $\mathcal{D}$.

5. **Case 1: $\triangle ABC$ is Non-Obtuse:**
- Then $d(ABC) = r(ABC)$.
- Points $A, B, C, X$ are in convex position. WLOG, assume $X$ lies in the region formed by rays $AB$ and $AC$.
- Let $A'$ be the antipode of $A$ with respect to $(ABC)$.
- If both $\angle XBA$ and $\angle XCA$ are $>90^\circ$, then $AX > AA'$. This implies:
$$d(AXB) \ge r(AX) > r(AA') = d(ABC)$$
which contradicts the maximality of $d(ABC)$.
- So, at least one of the angles $\angle XBA$ or $\angle XCA$ is $\le 90^\circ$. WLOG, assume:
$$\angle XCA \le 90^\circ$$
- Now, $\triangle XCA$ is non-obtuse and:
$$\angle AXC < \angle ABC \le 90^\circ$$
- By the Sine Rule, it follows:
$$d(AXC) = r(AXC) > r(ABC) = d(ABC)$$
which is a contradiction.

6. **Case 2: $\triangle ABC$ is Obtuse:**
- WLOG, assume $\angle C \ge 90^\circ$. Then:
$$d(ABC) = r(AB)$$
- From the Lemma, it is not hard to see that:
$$d(ABX) \ge r(AB)$$
with equality iff $X \in (AB)$. Since $X$ lies outside $(AB)$, we directly obtain our desired contradiction.

This completes the proof of the problem. $\blacksquare$