Given a graph $G$, prove (by contradiction) that if $\delta(G) \ge \frac{n}{2}$, then $G$ is connected. There is a similar problem here, but I am looking to prove this statement by contradiction, and I am having difficulty understanding these solutions.
Here is what I've got so far.

Proof.
  Let $G$ be a graph with $n$ vertices such that $\delta(G) \ge n/2$.
  Suppose for the sake of contradiction that $G$ is disconnected. 
  Let H be the smallest component of G. 
  Then $|H| \le n/2$.

From here, I know my general direction - I need to determine how many vertices, at most, can be in $H$ - which should conflict with what we know about $\delta(G)$.
Can anyone show me how to move forward with this idea?

If $\delta(G)\geqslant\frac n2$ but $G$ is not connected, then each component of $G$ must have at least $\left\lceil\frac n2\right\rceil+1$ vertices. But this means $G$ has at least $n+2$ vertices, a contradiction.
