Visualizing Infinity discerning countable and uncountable This is rather a philosophical question. Although it uses topological notions, it isn't any precise mathematics, so maybe one cannot take it very seriously.
Sometimes I try to picture an infinite set of points. Assuming that the space $E$ in which my visual imagination takes place is a simply connected bounded subspace of the plane $\mathbb{R}^2$, I can only visualize at most uncountably many points. But since $E$ is separable, so is every subspace and I can't really explicitly imagine uncountably many points, e.g. having a circle in mind: I couldn't decide whether there only lie denumerably points on it or not. I could decide for myself that there are uncountably many, but this I would consider cheating.
But I sometimes try to explicity imagine a countable set, viz as discrete, which isn't possible for uncountable sets, again because $E$ is separable. Since it's also bounded such a subset cannot be closed therein, for otherwise it would be compact, and therefore finite. One example would be the set $\{\tfrac{1}{n};\; n\in\mathbb{N}\}$, but here again it's hard for me to tell if its visualization is indeed denumerable. This might be due to the feeling that visualized points have an actual "size" in my mind.
And it seems that any bounded countable discrete set must locally look like this particular set because it must have an accumulation point outside of it. Or maybe one can determine that such a visualization is in fact denumerable?
The problem for me seems to be the phenomenon of an accumulation point, and I feel that there's no hope if "too many" of such points are actually in the set. I just don't know how many "too many" exactly is. Maybe one is already too much. And maybe it's enough if there's any accumulation point at all, i.e. not even in the set. In that case, it would be impossible (if my argumentation isn't wrong or an assumption too strict).
So my question would be if this is the end of this discussion or not? Is there any, maybe completely alternative, way of visualizing a countable set knowing it isn't uncountable? I don't know, the question might be very weird (I feel a bit ashamed posting it), but maybe someone has a satisfactory answer.
 A: I'll try to answer the question of how to visualize a countable set $C$ in, e.g., $\mathbb{R}^2$, although I'm not sure if this was your main question.
If $C$ is dense then I'm not sure how to visualize it differently than the ambient space.  Instead let's consider the special case that $C$ is closed.  In this case the usual proof of the Cantor–Bendixson theorem (see https://en.wikipedia.org/wiki/Perfect_set_property) maybe gives a way to "visualize" $C$ in terms of countable discrete sets, which you suggested that you already can visualize.
The proof of this theorem shows that if we transfinitely iterate the process of removing the isolated points of a countable set $C$, taking intersections at limit stages, we end up with the empty set after countable many stages.  So running this process in reverse, it says that $C$ is "constructed" in countably many stages by starting with a countable discrete set, and at each successor stage, adding another countable discrete set whose limit points are the current set.  (Well, what happens at the limit steps if you go in reverse isn't really a "construction", but a fuzzy question gets a fuzzy answer.) 
Edit by questioner: This answer actually didn't satisfy me, but the discussion in the comments of this answer eventually convinced me to believe that one only can visualize (in the sense I meant) finite unions of bounded connected subspaces of the plane, why the answer to my question would be "no, there is no way". This is why I accepted the answer.
