Set "in between" integers and reals All infinite sets that I know of are either countably infinite, or uncountably infinite. This leads me to ask the following question:
Is there any infinite set S s.t. there is no bijection between S and $\mathbb{R}$ or $\mathbb{Z}$?
 A: Every set is finite or infinite. Why? Because we have a certain definition for "finite" and we define "infinite" as "not finite". Similarly, we have a definition for "countable", and we define "uncountable" as "not countable".
However, much like you have different sizes of infinity (at least two that you know of, but in fact there are many more), saying that a set is "not finite" doesn't reveal to you anything about its cardinality. Well, just that it isn't finite. And the same goes for uncountability. When we say that a set is uncountable we don't have any other information about its cardinality, other than the fact that it is not countable.
Usually,$^1$ cardinals are denoted by $\aleph$ numbers, and so we have $\aleph_0$ is the first infinite cardinal, $\aleph_1$ is the first uncountable cardinal, $\aleph_2$ is the third infinite cardinal, and the second uncountable cardinal, and so on. And we have an operation which given a cardinal, gives us its successor (i.e. the least cardinal larger than itself). And we also have cardinal exponentiation, for example we know to calculate that $$|\Bbb R|=|\mathcal P(\Bbb N)|=2^{\aleph_0}.$$
We cannot, however, from the usual axioms$^2$ of set theory (and certainly not when we work in naive set theory) prove conclusively what is the exact $\aleph$ number of $2^{\aleph_0}$ (and for that matter, not of any cardinal exponentiation). So it is possible to have $2^{\aleph_0}=\aleph_1$, in which case the real numbers have the smallest uncountable cardinality, but it is also possible to have $2^{\aleph_0}=\aleph_2$, or almost any other value (almost because we do know to rule out some possible values, but very little of them). The assertion $2^{\aleph_0}=\aleph_1$ is known as The Continuum Hypothesis.
Suppose now that we have that $2^{\aleph_0}=\aleph_2$, then it means that there is a subset of $\Bbb R$ of cardinality $\aleph_1$ (because $\aleph_1<\aleph_2$). This subset is uncountable, but it's smaller (in terms of cardinality) than the real numbers themselves.
To conclude this ramble, uncountable simply means "not countable" and there are many different uncountable cardinals. It is possible that $2^{\aleph_0}=\aleph_1$, in which case every uncountable subset of the real numbers have the same cardinality as the real numbers, but it is also possible to have $2^{\aleph_0}$ much larger, in which case there are uncountable sets of real numbers of different sizes. The Continuum Hypothesis cannot be proved nor disproved using the usual axioms of set theory, and in order to settle it we need to add further axioms to our system.
And of course, whatever $\aleph_\alpha=2^{\aleph_0}$, its successor, the cardinal $\aleph_{\alpha+1}$, is strictly larger.

Further reading.

*

*How do the terms “countable” and “uncountable” not assume the continuum hypothesis?

*Why is CH true if it cannot be proved?

*Why is the Continuum Hypothesis (not) true?

*is there a cardinality between the rational and the irrationals?

Footnotes.

*

*I'm writing usually because we usually assume the axiom of choice, which is equivalent (amongst other thing) to the assumption that every set can be well-ordered, so every cardinal is an $\aleph$ number. Without the axiom of choice the cardinals become even crazier, and we cannot even decide whether or not $2^{\aleph_0}$ is an $\aleph$ number.
We can in fact formulate the continuum hypothesis in several non-equivalent ways without the axiom of choice, see How to formulate continuum hypothesis without the axiom of choice? for more information.


*The usual axioms of set theory are taken as $\sf ZFC$ here. These axioms are quite strong and allow us to prove a lot of things, but not the exact value of the cardinality of the continuum. There are several natural extensions which do decide the cardinality of the continuum, and sometimes we will just add an explicit axiom saying that $2^{\aleph_0}$ equals such and such.
A: All sets are either countable or uncountable.
However, for instance, $S=\mathcal{P}(\mathbb{R})$ will satisfy your last question. It's well known that given any set $X$, the cardinality of its power set $\mathcal{P}(X)$ is strictly greater than the cardinality of $X$. Since $|\mathbb{Z}| < |\mathbb{R}| < |\mathcal{P}(\mathbb{R})|$, setting $S=\mathcal{P}(\mathbb{R})$ gives you a set which bijects with neither $\mathbb{Z}$ nor $\mathbb{R}$.
If you're looking for a set which is uncountable but smaller than $\mathbb{R}$, then you're asking for a solution to the continuum hypothesis, which is (provably) unprovable from the ZF(C) axioms of set theory.
Bear in mind 'uncountable' really means 'not countable'; it certainly doesn't necessarily mean 'at least as big as $\mathbb{R}$'.
