Then the computational extension of the theorem is as follows:

If a (turing complete) machine can decide whether or not any given input will halt or not, then said machine cannot decide the input comprising itself.

By the law of the excluded middle, this is a contradiction, and as such, any such machine cannot possibly exist.

I don't think this statement is quite true. Specifically, the "input comprising itself" bit - I don't know of a proof that shows that this is the case. If you're alluding to the diagonalisation argument, the result is slightly more complex.

Suppose that A is a Turing machine that given any machine T and input I, decides whether B halts on I.

Let C be a machine that, when given I, calls A with machine I and input I and halts iff A concludes that I doesn't halt on input I.

If A is called with machine C and input C and halts, then that means that A must halt on C when given C. Inversely if A is called with (C, C) and doesn't halt, then that means A must halt on (C, C). Here lies the contradiction. (unless I've made a mistake)

So C is the input that breaks T, not T. Unless I've misunderstood what you've said.

But the more important question is - who are you? What kind of high school kid has this kind of background in maths? Where did you learn this stuff?