The set you are talking about is sometimes written as $\def\N{\Bbb N}\binom\N2$ in combinatorics: the set of all $2$-element subsets of$~\Bbb N$. It has the same cardinality as$~\N$. An explicit bijection is given by the combinatorial number system system for $k=2$: map the subset $\{a,b\}$ with $a<b$ to $\binom a1+\binom b2=a+\frac{b(b-1)}2$.
A similar bijection holds for $k$-element subsets for any fixed finite$~k$ (see the linked page). Even the union of all those sets is in bijection with$~\N$; in this case a slightly different bijection works: interpret the subset as a binary number, with the bit in position$~i$ indicating whether $i$ is present in the subset (value $1$) or not (value $0$).
In order to get an uncountable collection of subsets of$~\N$, your collection must contain some subsets that are infinite, and moreover need infinite information to describe (a necessary but not sufficient condition). As long as each element can be specified, using some fixed and definite encoding, by a finite amount of data, one can order the elements by size of those data first, and then for a fixed size by some total ordering of the encoding data (lexicographically for instance); this produces a well-ordering in which each element occurs in a position corresponding to a natural number. The above bijections are in fact inspired by fairly simple such ordering schemes, but using "largest element" in the place of "data size" (but that would no longer work for collections containing infinite subsets with finite descriptions).
There is an inherent problem with your question. Actually two.
The first is that visualizing something requires some sort of inherent structure. We can visualize the real numbers as a line, or the natural numbers as dots, and $\Bbb R^3$ as the room you are sitting in.
When you talk about cardinality, you—by definition—omit the structure. Note that $\Bbb N$ and $\Bbb Z$ are very different, but they are both countable. Even $\Bbb N$ with the usual order, and with the divisibility relation ($m\mid n\iff\exists k\in\Bbb N:mk=n$) are very different visually, despite being two structures of the same set.
So can we come up with an easy to visualize structure? Sure. We have linear orders of any cardinality, as follows from the axiom of choice, so we can take one which "locally" looks like $\Bbb R$, or even like $\Bbb N$ (again, locally is a key word here). Does that mean something to you? How is that different from $\Bbb R^2$ viewed with the lexicographic order? It really doesn't.
The second problem is relying on "visualizing". Yes, you can visualize the function $f(x)=x$ or $f(x)=e^x$. If you try very hard, you can even differentiate in your mind's eye between $\ln x$ and $\sqrt x$ (but I can't, their graphs are just too similar). Mathematics is built on definitions. You could argue, and correctly so, that $\Bbb Q$ and $\Bbb{R\setminus Q}$ are extremely different objects. While both are linearly ordered metric spaces, only one of them is completely metrizable, only one of them is a $G_\delta$ subset of $\Bbb R$, only one of them is uncountable, and only one of them is a field. Even though, they are both totally disconnected and without isolated points.
So if you think about either $\Bbb Q$ or $\Bbb{R\setminus Q}$ visually, you are likely to think about a line with "missing dots almost everywhere", but also dots are still everywhere". Sort of to signify that both the set and its complement in the real line are dense and co-dense sets. And therein lies the rub. We just talked about how different these mathematical objects are, from several important and distinctive perspectives. Yet, visually speaking, they "look and feel the same".
This is why visualizing is not a good direction. It is not a good motivating tool. Abstract mathematics, and set theory in particular, is prone to confuse you if you try to visualize things as a crutch and a scaffolding device. You should rely on the definitions you have, and work carefully. With time you will gain the proper intuition for handling definitions related to infinite sets and their cardinality. It's not the easy way, but you won't be cheating yourself with the idea that you can somehow visualize well enough to make the distinction between two objects which are very different from one another.
Best Answer
My favorite group of cardinality greater than the continuum is the group of field automorphisms of the complex numbers.
This is a very large and interesting group: not only does it have cardinality $2^{\mathfrak{c}} = 2^{2^{\aleph_0}}$ (to the AC patrol: yes, I'm assuming the Axiom of Choice here), it has this many conjugacy classes of involutions, which follows from (the affirmative answer I received to) this MO question. (This is also climbing my list of most frequently made errors by veteran mathematicians. I have watched many smart people claim that it follows from the Artin-Schreier Theorem that every index $2$ subfield of $\mathbb{C}$ is isomorphic to $\mathbb{R}$. The cardinal number by which this statement is off is pretty staggering.)
It's sort of a less fun answer, but: there are certainly groups of every infinite cardinality. If you want to be slick about it, this follows from the Lowenheim-Skolem Theorem in model theory (to the AC patrol...), since the theory of groups has a countable language and admits infinite models. One example is that the free abelian group (and also the free group) on an infinite set $S$ has cardinality equal to that of $S$. Moreover, the group $\operatorname{Sym} S$ of all bijections on an infinite set $S$ has cardinality $2^{\# S} > \# S$.
You can go on to construct your own favorite examples. E.g. there is a field $F$ of every infinite cardinality $\kappa$ (e.g. a rational function field over $\mathbb{Q}$ in $\kappa$ indeterminates) and for all $n \geq 2$, the group $\operatorname{PSL}_n(F) = \operatorname{SL}_n(F)/\text{center}$ of $n \times n$ matrices in $F$ with determinant $1$ modulo scalar matrices with determinant $1$ is a simple group of cardinality $\# F$. (This is a fun example because we also know all possible orders finite simple groups...but that's just a little bit harder!) All of these are fairly natural examples, I think.