What's the question? Is it the "challenge", and is this something you already know the answer to?

Ha! Hahaha! Oh, that anti- category theory humor just kills!

Right, I'm downvoted because why? This was a reference request.

Todd and Andreas: this is all interesting, and thanks, but mainly I was trying to clarify the intent of the question (and so my comment was directed to the OP).

By "quotable", I'm guessing you mean something you could easily say to a friend while on a hike.

Elizeu, that doesn't answer my question. If you have an example, please say what it is.

You say in general not. I don't think I follow. What example do you have in mind? (FWIW, Wikipedia seems to agree with me, en.wikipedia.org/wiki/Topological_group#Properties, paragraph 5, but it seems to me a straightforward argument.)

Where you say more generally, I assume you mean a subgroup equipped with the subspace topology? Because it seems to me the answer is yes, and it has nothing to do with total disconnectedness: a subgroup of a topological group, when given the subspace topology, is a topological group. Mostly it comes down to checking that if $H \hookrightarrow G$ is a subspace, then the product topology on $H \times H$ coincides with the subspace topology on $H \times H$ inherited from the product space $G \times G$.

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@Adrian , the OP doesn't visit MO regularly any more.

I must confess that it's not obvious to me that it's unbounded for any irrational $x$. Can someone set me straight?