Deciding whether set of running times is infinite
Suppose $L \in \textrm{RE}$.
Given a Turing machine $M$ you can construct another Turing machine $M'$ that takes an input $x$ and simulates $M$ on an empty input for at most $|x|$ steps. 

If $M$ eventually halts, say after $c$ steps, $M'$ will always halt after at most $f(c)$ steps, (for some function $f$), regardless of the choice of $x$. This shows that $\textrm{Count}(M') \subseteq \{ 1, \dots, f(c) \}$ is a finite set.

If $M$ does not halt, then $M'$ will halt after simulating $|x|$ steps of $M$ for every choice of $x$. By picking different values of $x$ it is possible to build an infinite set of inputs such that each input results in a different running time of $M'$ (e.g, start with $x=$"1" and let $s$ be the number of steps taken by $M'$ on input $x$. The next value of $x$ will be a string of $s+1$ ones, so that $M'$ necessarily halts after some number $\ell$ of steps such that $\ell \ge s+1$. The next value of $x$ will have $\ell+1$ ones, and so on).

This shows that $\textrm{Count}(M')$ is an infinite set.

Therefore, since $L \in \textrm{RE}$ you have just found a way to accept all (and only) the Turing machines $M$ that do not halt on an empty input, showing that the complement of the halting problem is in $\textrm{RE}$. 
This is a contradiction as it would imply that the halting problem is in $\textrm{R}$.