TITLE: Is there a theorem whose only known proof uses "$A$ or not $A$" for undecidable $A$
QUESTION [16 upvotes]: Is there a known theorem $T$ in $ZF+DC$ (or $ZF$ or $ZFC$) such that the only proof we know of $T$ is by using the LEM applied to $A$ ( "$A$ or not $A$" ), where $A$ is independent of $ZF+DC$ ?

REPLY [2 votes]: When I was in grad school I learned about the following fact which was actually intended as a joke.  
Theorem: There are irrational numbers $x$ and $y$ such that $x^y$ is rational.
Proof.  Look at $\sqrt2^{\sqrt2}$.  Is it rational?  If so, take $x=y=\sqrt2$, and we are done.  Otherwise, take $x=\sqrt2^{\sqrt2}$ and $y={\sqrt2}$, and we are again done since 
  $$x^y = \left(\sqrt2^{\sqrt2}\right)^{\sqrt2} = \sqrt2^{\left(\sqrt2\times\sqrt2\right)} = \sqrt2^2 = 2.$$
QED
I have no idea if the rationality of $\sqrt2^{\sqrt2}$ is decidable or not (most likely it is), so this is not quite an answer.