How to prove that if p $\le$ q+1 , then G is connected. prove if p $\le$ q+1 , G is connected. 
p is the number of vertices, q is the number of edges.
I would like to start with p=q+1. I remember that there is a theorem states "If G is connected, and p=q+1, then G is a tree." but I cannot use it because there is an assumption of G is connected. 
So I do not know how to start.
Here is the original question:
I simplify it to : 
2q $\ge$ 2(p-1)
q $\ge$ p-1
p $\le$ q+1
 A: Your first claim is wrong and this is a counter example.

Your second claim (edited version) is also wrong and this is a counter example.

A: Let $G$ be the disjoint union of two copies of the complete graph $K_n$. Then $G$ has $p=2n$ vertices and $q=n^2+n\gg p$ edges while not being connected.
A: Consider a graph $G$ that is disconnected and has a cycle. I claim we can produce another graph $G'$ with the same degree sequence as $G$ but with fewer connected components. Iterating, we end up with a graph with the same degree sequence that is either connected or acyclic. But an acylic graph satisfies $p=q-k$ where $k$ is the number of components; if $k>1$ this would contradict the assumption $p\geq q-1$.
To prove the claim, consider an edge $uv$ in a cycle, and any edge $xy$ in a different connected component. Construct $G'$ from $G$ by a "switch": deleting the edges $uv,xy$ and adding edges $ux,vy$. This doesn't change the degree sequence. Since $uv$ was in a cycle in $G$, $u$ is still in the same connected component as $v$ in $G'$, which implies that $u,x,v,y$ are all in the same connected component in $G'$. So any vertices that were connected before are still connected, and since $u$ and $x$ are now in the same component, the number of connected components has decreased by one as required.

Another argument is to inspect the construction in the proof of Erdős–Gallai as in the Havel–Hakimi algorithm, but that relies on knowing a particular proof.
