Consider the diameter graph $\Gamma$ on $n$ pairwise distinct lattice points in the plane, i.e., the geometric graph on those points, whose edges are the diameters of the configuration. We will prove that $\Gamma$ has at least one vertex of degree 1. Removal of one such and the edge it is incident to, allows then an inductive approach — the base case, $n=2$, is clear. As expected, the argument hinges on the fact that every two diameters intersect: Either they both emanate from the same point or they both cross at some interior point. If $\Gamma$ has a vertex $A$ of degree at least 3, proceed along the standard lines in the continuous realm: The diameters emanating from $A$ all lie within the angle formed by two diameters $AB$ and $AC$, where $\angle BAC \le 60^\circ$. Letting $AX$ be a third diameter from $A$, it then follows that $X$ is the desired vertex of degree 1, as a hypothetical diameter $XY$, $Y \ne A$, cannot intersect both $AB$ and $AC$.

It is a fact that $\Gamma$ has at most $n$ edges. Suppose, if possible, $\Gamma$ has $n$ edges and is 2-regular (all vertices have degree 2). As every two diameters intersect, this is possible if and only if $n$ is odd, in which case $\Gamma$ is a star-shaped self-crossing $n$-cycle $A_1A_2A_3\dots A_n$ whose polygonal convex hull reads around the boundary $A_1A_3A_5\dots A_nA_2A_4A_6\dots A_{n-1}$. Write $A_iA_{i+1} = d$, $i = 1, 2, \dots, n$ (indices are reduced modulo $n$), so $(x_{i+1} - x_i)^2 + (y_{i+1} - y_i)^2 = d^2$, $i = 1, 2, \dots, n$; clearly, $d^2$ is an integer. Without loss of generality, assume the configuration is the smallest possible.

By minimality, the $2n$ differences $x_{i+1} - x_i$ and $y_{i+1} - y_i$ cannot all be even — otherwise, the midpoints of all line segments $A_iA_j$ would again be lattice points, and a homothety of factor $\frac{1}{2}$ from some $A_i$ would yield a smaller configuration. Hence one of the index sets $I = \{i: x_{i+1} - x_i \equiv 1 \pmod 2\}$ and $J = \{i: y_{i+1} - y_i \equiv 1 \pmod 2\}$, say, $I$, is non-empty. Hence again, $d^2 \equiv 1 \pmod 4$, so for no index $i$ are $x_{i+1} - x_i$ and $y_{i+1} - y_i$ both even. Consequently, the union $I \cup J$ exhausts all $n$ indices.

Rule out the case $d^2 \equiv 1 \pmod 4$ as follows: If $d^2 \equiv 1 \pmod 4$, then $I$ and $J$ are disjoint. By the preceding, $I \cup J$ covers all $n$ indices, so $|I| + |J| = n$ which is odd. On the other hand, $\sum_{i=1}^n (x_{i+1} - x_i) = 0$ implies that $|I|$ is even; similarly, $\sum_{i=1}^n (y_{i+1} - y_i) = 0$ implies that $|J|$ is even, and hence so is $|I| + |J|$, contradicting the fact that $|I| + |J| = n$ is odd.

Finally, rule out the case $d^2 \equiv 2 \pmod 4$: If $d^2 \equiv 2 \pmod 4$, then $I = J$, so $|I| = n$, contradicting again the parity of $|I|$ established above. Consequently, the diameter graph $\Gamma$ is not 2-regular, as desired.

Remark. The required diameter upper bound is always achieved, so $n-1$ is, in fact, the maximal number of diameters a planar configuration of $n$ lattice points may have. To prove it, consider the Diophantine equation $x^2 + y^2 = 5^N$ $(*)$ where $N$ is a non-negative integer. By Jacobi's two-square theorem, the number of pairwise distinct solutions of the Diophantine equation $x^2 + y^2 = M$, where $M$ is a positive integer, is equal to four times the excess of the number of divisors of $M$ that are congruent to 1 modulo 4 over those congruent to 3 modulo 4. In the case at hand, $(*)$ has exactly $4(N+1)$ pairwise distinct solutions. This can equally well be established directly, by noticing that $1 \pm 2\sqrt{-1}$ are Gaussian primes, and $\frac{1}{\pi}\arg(1+2\sqrt{-1})$ is irrational; explicitly, the solutions are $u(1+2\sqrt{-1})^k(1-2\sqrt{-1})^{N-k}$, where $u = \pm 1$ or $\pm\sqrt{-1}$, and $k = 0, 1, \dots, N$. Exactly $N+1$ of these solutions lie in the first quadrant, $x > 0$ and $y \ge 0$; and since 5 is odd, exactly $\lfloor \frac{1}{2}(N+1) \rfloor$ of these, say, $(x_i, y_i)$, $i = 1, 2, \dots, \lfloor \frac{1}{2}(N+1) \rfloor$, satisfy $x > y \ge 0$. The $\lfloor \frac{1}{2}(N+1) \rfloor$ lattice points $(x_i, y_i)$ are all exactly $5^{N/2}$ away from the origin, and every two are (strictly) less than $5^{N/2}$ distance apart. Consequently, the origin and the $(x_i, y_i)$ form a planar configuration of $\lfloor \frac{1}{2}(N+3) \rfloor$ lattice points with exactly $\lfloor \frac{1}{2}(N+1) \rfloor$ diameters of length $5^{N/2}$ each. Setting $N = 2n-3$ completes the argument.

We end by describing a related configuration. Consider an even integer $N \ge n$. The $(x_i, y_i)$ above and the $(5^{N/2} - x_i, y_i)$ form a configuration of $N+2$ lattice points with exactly $N+1$ diameters: $\frac{1}{2}N+1$ of these join $(0,0)$ to each $(x_i, y_i)$, and another $\frac{1}{2}N$ join $(5^{N/2}, 0)$ to each $(5^{N/2} - x_i, y_i)$ with a positive $y_i$. Deletion of any $N-n+2$ points with both coordinates positive then settles the case.