Let $M$ be a model, $S$ assignment and $\varphi$ a formula, I need to prove, or disprove $M\vDash_S\exists x \varphi \rightarrow M\vDash_S \varphi$.
From what I understand $M\vDash_S\exists x \varphi$ means that there is an object $a$ in $M$'s domain such that for the assignment $S<x|a>$, $\varphi$ is true. Wouldn't that mean that $M\vDash_S \varphi$ is also true? Or could $S$ assign a different value to $x$ when it's a free variable in $\varphi$ but when $x$ is bound the object that $S$ assign doesn't matter because the quantifier $\exists$ just look for that one object in the domain? i.e $S$ can assign $b$ to $x$ so $M\nvDash_S \varphi$ but $M\vDash_S\exists x \varphi$ will still be true?
That's exactly right. $S$ assigns some value to $x$, but that value only matters when $x$ is a free variable--bound instances of $x$ completely ignore this assignment and instead are interpreted using the quantifier on $x$. So although $M\models_S\exists x\varphi$, there's no reason to think that the value $S$ assigns to $x$ is one that makes $\varphi$ true.