 Solution

Let $v_1\dots v_k$ be a longest path in the finite simple graph. Every neighbor of either endpoint lies on this path, since an outside neighbor would extend it. Define two subsets of $\{1,\dots ,k-1\}$:Their sizes are $d(v_1)$ and $d(v_k)$. The degree hypothesis gives $|I|+|J|\ge n\ge k$, but there are only $k-1$ possible indices. Hence some $i$ belongs to both. The edges at that index close a cycle through every vertex of the path:The graph is connected: any component has at least $n/2+1$ vertices by the minimum degree, so two components would exceed $n$ vertices. If $k<n$, connectedness gives an edge from this cycle to an outside vertex. Starting at that vertex and traversing the cycle without its closing edge gives a path with $k+1$ vertices, a contradiction.

Thus $k=n$ and the cycle is a Hamilton cycle, proving Dirac's theorem. The overlap-of-indices argument is why the degree threshold is half the number of vertices.