I found this proof on http://jeremykun.com/2012/02/08/busy-beavers-and-the-quest-for-big-numbers/ and have highlighted the part I don't understand in bold.
(BB(n) is defined as the number of steps made by n-state Busy Beavers.)
Theorem: BB(n) is uncomputable. That is, there is no Turing machine that can take as input n and computes BB(n).
Proof: Suppose to the contrary that such a machine existed, and call it T. We will use T to construct a machine which solves the halting problem as follows:
On input < T, w >, a description of a Turing machine and its associated input, we can determine in finite time the number of states that T has (it is encoded in the description of T).
Now simply compute BB(n), where n is the number of states in T, and simulate T on input w, counting its steps as it proceeds. Eventually the simulation of T either halts, or makes more steps than BB(n).
In the former case, we may stop the computation and declare that T halts. In the latter, we know that BB(n) is the maximal number of steps that can occur for any Turing machine with n states that eventually halts. Therefore, if T had not halted on the input w, it must never halt.
We may stop our computation there, proclaim an infinite loop, and thereby solve the halting problem.
This is a contradiction, so the sequence BB(n) cannot be computable.
If we assume that T can calculate BB(n) for any n, don't we also have to assume that it does halt, and therefore that it will not exceed the steps of BB(n)? (That would also mean that using T to solve the halting problem does not work)