1. Introduction and General Statement:
We aim to prove that for every positive integer $m$, there exists a finite set $S$ of points in the plane such that each point in $S$ has exactly $m$ points in $S$ at a unit distance from it. To do this, we will use a more general statement about unit distance graphs.

2. Unit Distance Graphs:
Consider the unit distance graph $G$ where $V(G) = \mathbb{R}^2$ and two vertices are adjacent if and only if the Euclidean distance between them is 1. We will show that if $G$ and $H$ are unit distance graphs, then their Cartesian product $G \times H$ is also a unit distance graph.

3. Cartesian Product of Graphs:
The Cartesian product $G \times H$ is defined as follows:
$$V(G \times H) = V(G) \times V(H)$$
and two vertices $(v_1, w_1)$ and $(v_2, w_2)$ are adjacent if and only if:
$$(v_1 = v_2 \text{ and } w_1 \leftrightarrow w_2) \text{ or } (v_1 \leftrightarrow v_2 \text{ and } w_1 = w_2)$$
Here, $\leftrightarrow$ denotes adjacency in the respective graphs.

4. Embedding Using Complex Numbers:
Suppose $G$ is embedded in the complex plane with vertices $v_1, v_2, \ldots, v_n$ and $H$ with vertices $w_1, w_2, \ldots, w_m$. We claim that for some choice of $0 \leq \theta < 2\pi$, the points $v_i + e^{i\theta} w_j$ will form the required embedding.

5. Condition for Adjacency:
We need to show that:
$$(v_i + e^{i\theta} w_j) \leftrightarrow (v_k + e^{i\theta} w_l) \iff (v_i = v_k \text{ and } |w_j - w_l| = 1) \text{ or } (w_j = w_l \text{ and } |v_i - v_k| = 1)$$
If the condition holds, adjacency is satisfied. Suppose $i \neq k$ and $j \neq l$, and the corresponding complex numbers are at a distance 1 from each other. This gives a quadratic equation in $e^{i\theta}$, which can have only finitely many solutions.

6. **Choosing $\theta$**:
By ruling out finitely many values of $\theta$ for each set of $i, j, k, l$, we can find a suitable $\theta$ that satisfies the adjacency condition.

7. Constructing the Graph:
For the given problem, we need a unit distance graph that is regular of degree $m$. We can start with the graph $K_2$ (a single edge) and form the $n$-dimensional hypercube $Q_n$ by iteratively taking the Cartesian product $Q_{n-1} \times K_2$. The $n$-dimensional hypercube $Q_n$ is a regular graph of degree $n$.

8. Conclusion:
By choosing $n = m$, we obtain a finite set $S$ of points in the plane such that each point has exactly $m$ points at a unit distance from it.

$\blacksquare$