I know that this is a repeat of a previous question asked with a similar title, but I didn't want to revive an old thread.
The solution presented in that thread seems to be the common one, but I was wondering if the alternate I came up with holds up and satisfies the claim.
If $G$ is a graph on $n$ nodes, where $n$ is an even number, then if every node of $G$ has a degree of at least $n/2$ then $G$ is connected.
Assume that $G$ is connected. Since it is connected, then by definition there exists a path between any two vertices, and there must be at least $n = 2$ vertices in $G$. Each vertex in $G$ has a degree of at least one.
Adding edges to a graph that is already connected (in order to satisfy the requirement that every node has a degree of at least $n/2$) does not destroy its connectivity, and so the claim is true.
I'm assuming that n can only be positive. The examples I drew to support my proof all look like singly-linked linked lists, eg 1-2-3-4-5-6 (undirected, of course).
Besides not being sure if the logic holds, I'm also not sure if it's acceptable to start by assuming that what we're supposed to be proving is true and then working backwards from that.