I wanted to prove that if $F$ is a field and we consider the fraction field $F(x)$, then $\sqrt{x} \notin F(x)$.
I said that if $\sqrt{x} \in F(x)$ then there are $f,g \in F[x]$ such that $\bigl(\frac{f(x)}{g(x)}\bigr)^2=x$ so $f(x)^2=xg(x)^2$ so $2(\deg(f)-\deg(g))=1$ which is not possible. Is this true? Is there a quicker way to see this?
That's correct and I can't see a quicker way. You should observe that $f(x)\ne0$ and $g(x)\ne0$, but otherwise it's fine.
Probably I'd write $$ 2\deg f(x)=1+2\deg g(x) $$ and observe that the left-hand side is even, while the right-hand side is odd, but it's just personal preference.
By the way, if one defines a “degree” in $\mathbb{Z}$ by declaring that $\deg n$ is, for $n\ne0$, the number of prime factors in $n$, so $\deg 1=0$, $\deg 2=1$ and $\deg12=3$; this function is well defined because of uniqueness of factorization and $\deg(mn)=\deg m+\deg n$. Using this degree, irrationality of $\sqrt{2}$ is proved in exactly the same way; more generally, this proves the irrationality of $\sqrt{p}$ for every prime $p$. Using the “$p$-degree”, defined in an obvious way, one can prove the irrationality of $\sqrt{d}$ for every square free positive integer $d$.