Consider a set of $N$ points $P=\{p_1, p_2, \dots p_N \}$ on $\mathbb{R}^2$ such that the distance $D$ between any two pair of points is less than or equal to 2:
$$D(p_i, p_j) \le 2 \qquad \forall \, i,j \in {1,\dots,N}. $$
That is, the points $p_i$ have unit circles around each point, and each unit circle $A_i$ overlaps with the others:
Question
Can it be proven that the region $R$ is not empty, where $R$ is the region of all points $a$ that also have a distance $D \leq 1$ to all points in $P$? E.g., $$ R = \{ a \in \mathbb{R}^2 : D(a, p_i) \leq 1 \qquad \forall \, i\in \{1,\dots,N\} \} $$
Method so far:
My approach so far has been to define the set of points $A_i$ as the set of points with distance $D\leq1$ to point $p_i$. Then the condition of pairwise distances can be written as the intersection of two sets: $$ A_i \cap A_j \ne \phi \qquad \forall \, i,j \in {1,\dots,N}, $$ where $\phi$ is the empty set. The region $R$ is then the intersection of all sets $A_i$: $$ A_1 \cap A_2 \cap \cdots \cap A_N = \bigcap^N_{i=1} A_i = R \neq \phi $$
I would proceed with a proof by induction on $N=1$ and showing $N \to N+1$ but introductory math was a long time ago so I am having trouble finishing the proof. Any help is appreciated.
Consider the case where the $p_i$ are the three vertices of an equilateral triangle of sidelength $2$. The disks of radius $1$ about each vertex intersect with each other in just the three midpoints of each side, and obviously the overall intersection is empty. You can put any number of additional $p_i$ inside the triangle near each of the vertices without their disks reaching the center of the triangle where any mutual intersection would occur. $R$ can easily be empty.