TITLE: Why does creating a model show consistency?
QUESTION [12 upvotes]: As per the title, why does the ability to generate a model from axioms prove they are consistent?

REPLY [5 votes]: I would just like to mention that one generally does not construct a model of a theory $T$ from the axioms of $T$ itself.  Instead, one uses external means to produce a structure that one can show is a model of $T$.
Here is where Ben's answer comes into play, although in my opinion it obfuscates matters: if you construct a model of $T$ using some other theory $T^\prime$, then what you have really done is shown that the consistency of $T^\prime$ implies the consistency of $T$.  There are times where this distinction is important.  For example, from the axioms ZFC of set theory one can construct a model of Peano Arithmetic, PA.  It is unknown whether ZFC is actually consistent, though widely believed, and therefore we cannot immediately conclude that PA is consistent, though this is also widely believed.
But let's assume that we have constructed a model $M$ of $T$ using methods that are above reproach.  For example, $M$ might be a finite $\mathcal{L}$-structure of specific size $n$.  The claim that $T$ is consistent then comes down to two facts about the underlying logic.
Fact 1: If $M$ is a structure in some language $\mathcal{L}$, then $M \models \varphi$ and $M \models \neg \varphi$ cannot both be true for every $\mathcal{L}$-sentence $\varphi$.  (In fact, exactly one of $M \models \varphi$ or $M \models \neg \varphi$ holds.)
Fact 2: If $T$ is a theory in a language $\mathcal{L}$ and $\varphi$ is a $\mathcal{L}$-sentence such that $T \vdash \varphi$, then $M \models \varphi$ for all models $M$ of $T$.
Now, if $\varphi$ were an $\mathcal{L}$-sentence such that $T \vdash \varphi$ and $T \vdash \neg \varphi$, then by Fact 2 we conclude that $M \models \varphi$ and $M \models \neg \varphi$, but this contradicts Fact 1!