Show that "$\Gamma \models S \Rightarrow \Gamma \vdash S$" entails "if $\Gamma \nvdash P \And \sim P$ then $\Gamma$ is satisfiable" 
Show that "$\Gamma \models S \Rightarrow \Gamma \vdash S$" entails "if
  $\Gamma \nvdash P \And \sim P$ then $\Gamma$ is satisfiable"

I'm primarily confused with the notation being used here. In particular, I understand that "satisfiable" means there is at least one interpretation that makes $\Gamma$ true, but I'm not sure what $\vdash$ means nor what the statement $\Gamma \models S \Rightarrow \Gamma \vdash S$ means. The first part tells me that $\Gamma$ "makes true" $S$, or that there is no interpretation which makes $\Gamma$ true and $S$ false, but I don't understand how to interpret what comes after this (the "$\Rightarrow \Gamma \vdash S$" part). I'd really appreciate if someone could clear up this notation for me.

So I think this is what we may want to do.  Suppose for contradiction that $\Gamma$ is not satisfiable.  This means that $\Gamma$ has no models.  Now, fix some sentence $P$ and let $S \equiv P \wedge \neg P$.  Now, $\Gamma \models S$ will be vacuously true since there are no models of $\Gamma$, (i.e. any model of $\Gamma$ will satisfy $S$).  So by the hypothesis, we have that $\Gamma \vdash S$.  But this means $\Gamma \vdash P \wedge \neg P$ contradicting the hypothesis that this does not happen.
