Candidates for uncountable subsets of $\mathbb{R}$?

Maybe this is close to your thinking:

Take a set $A\subset \mathbb R$ that is uncountable and take any subset $B\subset A$ that is countable. Then the set $A\setminus B$ is uncountable. Do all examples of uncountable sets look much like this? Is this the only method of finding such examples?

So for example, we know an interval $[a,b]$ is uncountable. So a simple example of a relatively interesting uncountable set is $[a,b]\setminus S$ where $S$ is some set that I can reasonably show is countable (e.g., all rationals, all algebraic numbers, the range of some sequence, etc.)

But there are much more interesting examples of uncountable sets. Every perfect subset of $\mathbb R$ (i.e., every nonempty closed set without isolated points) is uncountable and in fact has cardinality $\mathfrak c$ of the continuum. All of the Cantor-like sets (including the orginal Cantor ternary set) are perfect and so uncountable.

For this see: Proof that a perfect set is uncountable

If you have studied Lebesgue measure on the real line then you have an abundance of interesting uncountable sets. If $E\subset \mathbb R$ is a measurable set of positive Lebesgue measure then it is trivial to show that $E$ is uncountable. This cheats a bit since every such set $E$ contains a perfect set so we are back to saying it is both uncountable and has cardinality $\mathfrak c$.

Of course there are plenty of other ways an uncountable set might arise naturally. For example let $f:[a,b]\to \mathbb R$ be a continuous function with the property that every level $$f^{-1}(L)= \{x\in [a,b]: f(x)=L\}$$ is countable (i.e., finite or countably infinite). Then there is an uncountable set of points where the derivative $f'(x)$ must exist.

It follows, too, that if $f$ has a derivative at only a countable number of points then there are uncountably many values of $L$ so that the levels $f^{-1}(L)$ are themselves uncountable. (By the way, this leads us back to perfect sets: levels of a continuous function are always closed and most continuous nowhere differentiable functions must have most levels perfect.)

In short you will, before long, see more instances of uncountable sets than you might care to confront.

There are lots of uncountable subsets of $\Bbb R$. There are $2^{\mathfrak c}$ of them because there are $2^{\mathfrak c}$ subsets of $\Bbb R$ and only $\mathfrak c$ of them are countable. You and I can only imagine a vanishingly small fraction of them. We have to prove things about them without imagining the variety of them.