1) There is no algorithm that can compute arbitrarily many digits of $\Omega$. If we were in some fashion capable of checking whether given strings of numbers composed the first $n$ of $\Omega$'s binary expansion, we could simply brute force our way through all numbers with denominators $2^k,$ $k=1,2,3,\dots$, which would provide us with more and more digits of $\Omega$ - a contradiction.
However, all is not lost. First off, the set of rational numbers less than $\Omega$ is computably enumerable, so there exists an algorithm that lists them off one by one. More directly, we can compute successively better lower bounds by simply calculating known halting programs and discovering more of them (we never know how close we are, though, or otherwise we could extract digits). If the aliens give us a number less than Chaitin's constant, then in theory we could confirm that. In practice, though, we might well not have enough time and resources to check their number if it's too close to the real value, so we wouldn't be sure (and of course, we would be guaranteed failure if their number is $>\Omega$).
Second, just because no algorithm computes infinitely many digits, doesn't mean we can't soundly prove the first $n$ digits for some fixed $n$ (and a fortuitous choice of Turing machine). Mathworld provides two examples of Chaitin's constant being computed for two TM's:
$$\Omega_U= 0.0000001000000100000110\cdots_2$$
$$\Omega_U=0.0001000000010000101001110111000011111010\cdots_2 $$
For more on these, you'll have to see Calude (2002) and Calude, Dinneen (2007). Using these, we can do a rudimentary check on the alien's number by confirming or disconfirming that it agrees with what we have. But of course, this is extremely limited - and highly dependent on choice of Turing machine.
Third, the digits of $\Omega$ are algorithmically random, i.e. the first $n$ digits (alone) can only be enumerated in at least $n-O(1)$ steps. Thus it might be possible to investigate the plausibility of candidate digits via approximations of Kolmogorov complexity or other statistical tests. However, this may require a very large number of candidate digits in order to obtain a useful indicator, possibly more than is feasible to realistically work with.
2) Whether the answer to a general mathematics problem is "of any use" is outside the scope of computability theory. To the extent that proving or disproving a conjecture is inherently useful, then calculating truth or falsity based on $\Omega$ would be useful. If an existence claim turned out false, we could stop searching computationally, or if a universality claim turned out false we could start searching computationally for a counterexample. Precisely what the implications of a conjecture are depend on the problem itself, e.g. the Riemann hypothesis affects the distribution of prime numbers and subsequent arithmetic bounds in number theory.
However, a simple computation of this sort won't lend any insight into why a problem turns out the way it does. Normally in mathematics, famous unsolved problems require whole new theories, perspectives, or directions to resolve, but we won't get that from a $\Omega$-based calculation because all of the "reasoning" is not found in the digits, nor the checking of the digits. Chaitin's constant encodes only the answers, not the why's behind them.
This is not what you asked for, because it's not a proof. But it is an argument.
Consider some famous and unresolved problem of mathematics, such as the twin primes conjecture. (Or the Collatz conjecture, the Goldbach conjecture, or, until recently, the Fermat conjecture.) I am sure you recall that the conjecture is that there are arbitrarily large numbers $p$ and $p+2$ that are both prime. ("Twin primes")
It's easy to write a computer program which, given an input $N$, looks for a pair of twin primes larger than $N$, and which halts when it finds such a pair.
The twin primes conjecture is true if, and only if, this program halts on all inputs.
Therefore, if there were a reliable way to tell if a program halts on all inputs, the twin primes conjecture would be easy to resolve, and we could conclude from the fact that it is not resolved that mathematicians are all a bunch of dummies.
Now maybe your students don't care about the twin primes conjecture and perhaps they are not sure that mathematicians are not all a bunch of dummies. But they are familiar with problems and puzzles in general, and they are probably familiar with the idea that it is sometimes not only hard to find solutions, but it can even be hard to see ahead of time if if there is a solution. They can probably be persuaded that you can get a computer to search exhaustively for the solution to some problem, and halt and print it out if it finds one.
A solution to the halting problem would mean that a very large class of problems could be easily solved. You would write a program to search for a solution, and then, instead of running it, you would analyze the program to see if it would halt. If the answer was yes, you would know there was a solution, and all you had to do to find it was to run the program. On the other hand if the program would not halt, you wouldn't have to bother to run it because you would know there was no solution.
That would be a very different world from the one we actually live in, and it may be possible to persuade the students of that.
Best Answer
Let me start by noting that if there were a simple algorithm to solve the halting problem, much of modern mathematics would be completely different. For instance, consider the following program:
Whether this program halts is equivalent to Goldbach's conjecture! So the halting problem is a pretty sweeping thing. So the reason that this problem is unsolvable is not that there is too little information encoded in it, but far too much.
The unsolvability of the halting problem means specifically that there is no Turing machine that, when presented with the input of another Turing machine (in any reasonable description) and some input, will determine whether the input Turing machine will halt on the input string (and always output a yes or no answer in a finite amount of time). This does not, of course, preclude people from finding specific ways to prove that specific Turing machines halt or do not halt, but means that there is no general way (at least, insofar as "way" means "method that a Turing machine can emulate") that will work for any such program.