If you are willing to take the completeness theorem for granted, then it can be quite easy.
If $T$ is a first-order theory, then $T\vdash\varphi$ if and only if $T\models\varphi$.
This means that in order to prove that $\varphi$ is unprovable from $T$, then we only need to exhibit a model of $T$ in which $\varphi$ is false.
For example, we can prove that as a field $\Bbb R\models\exists x(x\cdot x=1+1)$, but the Greek also proved that $\sqrt2$ is not a ratio of two integers, therefore $\Bbb Q\models\lnot(\exists x(x\cdot x=1+1))$. And so we proved that the theory of fields does not prove the existence of $\sqrt2$.
Andre suggested in the comments a similar example with group theory and the statement "multiplication is commutative".
Other even easier examples may include things like a language $\cal L$ with a single constant symbol $c$, and the empty theory. Then $\forall x(x=c)$ is not provable, since in any structure for $\cal L$ with more than one element $\exists x(x\neq c)$ is a true statement.
Or even without constants, just look at $\forall x\forall y(x=y)$.