Prove that a graph G is connected if and only if it has a spanning tree?

Proving this is simple. First, you prove that G has a spanning tree, it is connected, which is pretty obvious - a spanning tree itself is already a connected graph on the vertex set V(G), thus G which contains it as a spanning sub graph is obviously also connected.

Second, you prove that if G is connected, it has a spanning tree. If G is a tree itself, then it must "contain" a spanning tree. If G is connected and not a tree, then it must have at least one cycle. I don't know if you know this or not, but there is a theorem stating that an edge is a cut-edge if and only if it is on no cycle (a cut-edge is an edge such that if you take it out, the graph becomes disconnected). Thus, you can just keep taking out edges from cycles in G until all that is left are cut-gees. Since you did not take out any cut-edges, the graph is still connected; since all that is left are cut-edges, there are no cycles. A connected graph with no cycles is a tree. Thus, G contains a spanning tree.

Therefore, a graph G is connected if and only if it has a spanning tree!