Is L={|M is a TM and L(M) is uncountable} decidable? Is $L=\{\langle M\rangle\mid \text{$M$ is a Turing machine and $L(M)$ is uncountable}\}$ decidable? 

My intuition is that it is not, but I'm not sure if Rice's Theorem applies in this case. If it is not decidable, how can I prove that using reducibility?
This is somewhat of a trick question. What you are missing is that there are no uncountable languages over a finite (or even countable) alphabet. This should be enough information to answer it. 

(I initially gave the entire answer away, but edited it out after considering a bit.)