Proof that If G = (V , E ) is a graph, then S is an independent set ⇐⇒ V − S is a vertex cover. I have the following :
  Proof:
Proof.  ⇒ Suppose $S$ is an independent set, and let $ e = (u, v )$
be some edge. Only one of $u, v$ can be in S . Hence, at least one of
$u, v$ is in $V − S$ . So, $V − S$ is a vertex cover.
⇐ Suppose $V − S$ is a vertex cover, and let $u, v ∈ S$ . There
can’t be an edge between $u$ and $v$ (otherwise, that edge wouldn’t
be covered in $V − S$ ). So, S is an independent set. #
But when I construct a Graph as shown bellow:

I supposed the blue nodes to be the Independent Set IS. and after that I picked up an edge $e1= (5,6)$. but I see it does not satisfy the first condition of the proof "...and let $ e = (u, v )$
be some edge. Only one of $u, v$ can be in S ", Where is the mistake?

It's correct, but perhaps the wording is a bit vague/imprecise. A better way to say it is:

At most one of $u,v$ can be in $S$.

Then when we negate both sides, we see where the "at least one" part came from.
