For all real numbers in bulk— You may call it a diagonal argument, but it’s just a reduction to Cantor’s original statement, no new proof needed. There are only countably many computable numbers, because there are only countably many programs, because there are only countably many finite strings over any finite alphabet[1].

For individual real numbers— There are of course provably uncomputable ones. Chaitin’s constant is the poster child of these, but you could just take a map of (number of Turing machine in any numbering of all of them) to (terminates or not) and call that a binary fraction. (This is actually not far away from Chaitin’s constant, but the actual one is reweighted a bit to make it more meaningful.) Are there unprovably uncomputable ones? At a guess I’d say so, but I’m not good enough to give a construction offhand.

[1] A countable union of (finite or) countable sets is finite. Rahzrengr gur havba nf sbyybjf: svefg vgrz bs svefg frg; frpbaq vgrz bs svefg frg, svefg vgrz bs frpbaq frg; guveq vgrz bs svefg frg, frpbaq vgrz bs frpbaq frg, svefg vgrz bs guveq frg; rgp. Vg’f snveyl boivbhf gung guvf jbexf, ohg vs lbh jnag gb jevgr gur vairefr znccvat rkcyvpvgyl lbh pna qenj guvf nf n yvar guebhtu gur vagrtre cbvagf bs n dhnqenag.