I was reading a result where the following proposition appears as a preliminary step (and left as exercise):
Claim: Suppose $G$ is a graph on $n$ vertices ($n$ even and $n \geqslant 3$) with minimum degree at least $n/2$. Show that $G$ contains a perfect matching.
Proof: By Dirac's theorem, $G$ has a Hamiltonian cycle $C$. Since $|C|=n$ is an even integer, the set of “odd” edges of $C$ gives a perfect matching for $G$. $\quad\Box$
I feel that using Dirac's theorem for this claim is an overkill. But after trying for a few days, couldn't come up with any other proofs. Can you give a more “direct” proof of the claim that avoids the machinery of Hamiltonian cycles?
Naturally, you might object to the vague requirement of “directness”. To clarify what I mean by it, I’ll give an example.
Claim 2. Suppose $G$ has a minimum degree at least $n/2$. Show that $G$ is connected.
Proof via Dirac's theorem. $G$ has a hamiltonian cycle as before, and hence is connected. $\Box$
A different proof. We'll show that any pair of vertices $u,v$ are connected by a path of length at most $2$. If $uv$ is an edge, we are done. Suppose not. Then $N(u) \cup N(v) \subseteq V \smallsetminus \{u,v\}$. Therefore
$$
|N(u) \cap N(v)| = |N(u)| + |N(v)| – |N(u) \cup N(v)| \geqslant \frac{n}{2}+\frac{n}{2}-(n-2) = 2 \gt 0.
$$
In particular there exists vertex $w$ such that $uw$ and $wv$ are both edges, and we are done. $\quad \Box$
I find the proof via Dirac's theorem much less illuminating in this example. In fact, as Qioachu points out below, the Claim 2 might even appear as an intermediate steps of Dirac’s theorem, making the above proof cyclic. My intention here is only to point out that Dirac’s theorem can be used as a powerful black box in killing much easier results.
Best Answer
Here's an alternative proof using Hall's theorem -- I don't know whether that counts as a more "direct" proof, but at least it uses a theorem that directly deals with perfect matchings instead of a detour (pun intended) through Hamiltonian cycles.
Arbitrarily divide the vertices into two sets $A$ and $B$ of equal size. In each set, find a vertex with a minimal number of edges connecting it to the other set, and swap the two minimal vertices. Repeat this until the swap would no longer increase the total number of edges connecting the two sets. (Since there is a finite number of edges, this must happen at some point.) Denoting the two vertices whose swap would no longer increase the number of connections by $v_A$ and $v_B$ and the numbers of their edges within and between the sets by $v_{AA}$, $v_{AB}$, $v_{BA}$ and $v_{BB}$, and counting the number of connections between the sets before and after the swap, we have $v_{AB}+v_{BA}\ge v_{AA}+v_{BB}$. (There might be an edge between $v_A$ and $v_B$, but that works in favour of the inequality.) It follows that
$$v_{AB}+v_{BA}\ge \frac12(v_{AB}+v_{BA}+v_{AA}+v_{BB})=\frac12(\deg v_A+\deg v_B)\ge \frac12(n/2+n/2)=n/2\;.$$
Now consider a subset $X$ of $A$. If $|X|\le v_{AB}$, then since each element of $A$ has edges to at least $v_{AB}$ elements of $B$, $X$ has at least as many neighbours in $B$ as elements. If $|X|>v_{AB}$, then since each element of $B$ has edges to at least $v_{BA}$ elements of $A$ and $v_{AB}+v_{BA}\ge n/2=|A|$, at least one of these edges must lead to $X$, so $X$ has $n/2$ neighbours in $B$, and thus at least as many as it has elements. Thus the premise of Hall's theorem is fulfilled, and the bipartite graph induced on $A$ and $B$ must contain a perfect matching, which is also a perfect matching in $G$.