Square class of algebraic extension of finite fields This is Q17 in Chapter 2 of the book "Introduction of quadratic forms over fields".

Let $F$ be an algebraic extension of a finite field $\mathbb F$. Show that $|F^*/(F^*)^2| \le 2$. 

If the extension is finite, then $F$ is itself a finite field and indeed $|F^*/(F^*)^2| = 2$ (basically because $F^*$ is cyclic). 
An example will be 
$$ F = \bigcup _{n=1}^\infty \mathbb F_5\left(\sqrt[2^{\ n}]{2}\right),$$
where $|F^*/(F^*)^2|=1$ (also an exercise in the book). 
I don't even know if I could use any theory of quadratic form to tackle this. The closest one is of course the Pfister's Theorem, which says 
$$I(F)/I^2(F) \cong F^*/(F^*)^2,$$
where $I(F)$ is fundamental ideal in the Witt Ring $W(F)$. 


Corollary 3.6 Let $F=\mathbb{F}_q,$ ($q$=odd). 
   A) If $q\equiv 1(\mod 4)$, then $W(F)\cong \mathbb{Z}_2[\dot{F}/\dot{F}^2].$
   B) If $q\equiv 3(\mod 4)$, then $W(F)\cong \mathbb{Z}_4.$ 

Since all finite algebraic extensions of a finite field are finite, then the theorem applies, and  you can use $I/I^2\cong\dot{F}/\dot{F}^2$ to make the desired assertion since you know the Witt rings up to isomorphism.
On the other hand, if $F$ is the algebraic closure of some finite field $GF(p),$ for prime $p,$ then $\dot{F}/\dot{F}^2=1,$ and you have the desired claim.
For an infinite extension of $F/GF(p)$ for some prime $p,$ that is not the algebraic closure, you should consider that this will be the direct limit of a set of finite fields (like the union you mentioned), which produces a direct limit of the group of units. You can then make an argument about the square classes of each finite extension, and get a direct limit of the square class groups, and classify the square class group $\dot{F}/\dot{F}^2$ up to isomorphism (in this case either trivial or cyclic of order 2).
