Trying to describe this Turing Machine Let's say I have the following turing machine:
$F_n$ = {M | M is a TM and |L(M) ≤ n}
In english, for some given natural number n, $F_n$ is the language of all turing machines that accept no more than $n$ strings.  We want to prove that for all values of $n$, that $F_n$ is co-RE.
I need to describe a turing machine that is a co-recognizer for $F_S$, which means that it must reject all M if there exist more than $n$ strings that $M$ accepts.
 A: Let $E_n = \{M : M \text{ is a TM and } |L(M)| > n\}$. 
Note that $E_n$ is the complement of $F_n$. To show that $F_n$ is co-r.e., one needs to show that $E_n$ is r.e. 
As usual Turing machines are coded by natural numbers and Turing machines take input natural numbers. (All other alphabets can be coded as natural numbers.)
So a Turing machine $U$ that accepts $E_n$ is as follows: On input $e$ (which is the code of machine $T_e$), simulate $T_e$ first on input $0$ for $0$-steps, then simulate $T_e$ input $0$ and $1$ both for $1$ step, then input on $0$, $1$, $2$ all for $2$ steps, etc. Halt at step $s$ only if ever you find $n + 1$ numbers less than $s$ such that $T_e$ halts within $s$ steps. 
The machine described above will accept $E_n$. So $E_n$ is r.e. Hence, $F_n$ is co-r.e. 

I have never heard of a co-recognizer; however, from the definition you provided above, it would essentially be the same machine above except where I said halt, you would reject.  
A: Run all programs on all inputs and keep track of how many inputs each program has halted on so far.  Whenever a program has halted on $n+1$ inputs, output it.  This enumerates the complement of $F_n$ so $F_n$ is in co-RE.
