Proving a language as undecidable without using reductions Let's say our Σ is 0 and 1. I want to disprove the following:


  There can be Turing Machines that accept only 1's, i.e. 1, 11, 111,
  etc. Therefore, all languages that have strings of 1's are
  recognizable.


My first step to proving this is by providing a language such as $One$, that accepts only if $w$ contains only $1$'s. There is a theorem that stated that $One$ is decidable iff $One$ and $\overline{One}$ are recognizable.

So, if I were to prove either one of them as unrecognizable, then I should be good. $\overline{One}$ I'll define as the set that doesn't contain 1 in it (i.e. episolon, 0, 00, etc). Creating a turing machine that simply goes right and accepts if we reach end without hitting a one, rejects if we do hit a one seems to work for this case. Could a language be created that isn't recognizable?
It isn't recognizable because it is infinite, therefore it'll loop infinitely. Does this work?


No, because that claim is false. And your writing isn't at all clear, here. Recognizability is a property of languages, so your first two "it"s must refer to languages; but "loops infinitely" is a property of Turing machines, so your third "it" must be one of those.

Ultimately, though, your whole approach is doomed because the languages you call $One$ and $\overline{One}$ are both decidable. You can easily see that $One$ is decidabe just by designing a Turing machine that decides it.