Is $\{w~|~\forall x \in T(M_v):|w|>|x|~\}$ decidable?
Yes, you are missing something.
The first argument that this language is co-semi-decidable is correct. However, the second one is wrong (and not formally written. A formal proof could not consist of such arguments, they are only for intuition usually).
Now to show why it is fully decidable:
We know that $|T(M_v)|=c&lt;\infty$. Now, since this is a constant smaller than infinity, there must be an $x_0\in T(M_v)$ for which $|x_0|$ is the longest from the words in $T(M_v).$
Now, build the turing machine that accepts $w$ iff $|w|&gt;|x_0|$.
Since $x_0$ is the longest, then if $|w|&gt;|x_0|$ we will have that $\forall x\in T(M_v):|w|&gt;|x_0|\ge|x|$ and thus $w$ is in your language. Now, if $|w| \le |x_0|$, then obviously $w$ is not in your language.
Thus $w$ is in your language iff $w$ is accepted by the turing machine we built.