An argument that $0\lt\Bbb{R}\lt 1$ is actually a countably or unaccountably infinite set? I currently have upheld an argument with myself on this question for the longest time. Here is my argument for both sides. 
Countable: You can express $0\lt\Bbb{R}\lt 1$ like this: 
$[\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{3}{4}\cdots]$
And that all of the fractions tend incandescently close to a transcendental or irrational #. For example fractions generated would come closer and closer to $\pi-3$, which is transcendental. A few numbers in this set that would asymptotically converge on $\pi-3$ are the fractions $\frac{1}{7},\frac{15}{106},\frac{16}{115}\cdots\to\pi-3$ And because there are infinitely many parts to this all of the numbers would eventually (at infinite'th #) reach whatever constant you want.
Uncountable: Because of the definition of a transcendental #, no transcendental number can be created as a fraction, and thus is not part of the group of fractions. That a number cannot be expressed as a fraction means that you must be forced to add $\epsilon$ to reach every transendential, making the entire set become uncountable. (As the difference between $a$ and $a+\epsilon$ is infinitesimally small.)
I am getting a little agitated researching rules on my own and having to judge the mixed results I am getting. Please help me decipher the technicalities of the rules of this set problem please. Any help appreciated! I also would like to apoligize if my question makes no sense.
 A: Assuming your notation $0 < \mathbb R < 1$ means $\{x \in \mathbb R \mid 0 < x < 1\}$, this set is uncountable, and there are many ways to show it. Here is a variant of the Cantor argument which is a bit nicer because we don't have to talk explicitly about decimal expansions or whatnot. 
Suppose for a contradiction that $[0,1]$ is countable. Let $(x_n)$ be an enumeration of $[0,1]$.
Let $I = [0,1]$. Divide this interval into three subintervals of equal length $1/3$: namely, $[0,1/3]$, $[1/3, 2/3]$, and $[2/3, 1]$. Now $x_1$ cannot be in all three of these intervals (indeed, unless it is one of the endpoints, it is in exactly one of them). Choose $I_1$ to be one of these subintervals which does not contain $x_1$. Now divide $I_1$ into three subintervals of equal length $1/9$. Then $x_2$ cannot be in all three of these subintervals. Choose $I_2$ to be one of the subintervals that does not contain $x_2$.
Continue this process indefinitely. At stage $n$ we select an interval $I_n$ of length $1/3^n$ such that $x_n \not\in I_n$. Since each $I_n$ is a subinterval of $I_{n-1}$, the intervals are nested: $I_1 \supset I_2 \supset I_3 \supset \cdots$. Since the intersection of a a nested sequence of nonempty closed bounded sets is nonempty, there is a point in the intersection: $x \in \bigcap_{n=1}^{\infty} I_n$. Note that $x$ cannot be any of the $x_n$'s, because we chose each $x_n$ so that it is not in $I_n$, hence not in the intersection.
Thus we have shown that no enumeration of $[0,1]$ can include all of the points in $[0,1]$, so $[0,1]$ is uncountable.
(Of course, $(0,1)$ is also uncountable, for if it were countable, then $[0,1] = (0,1) \cup \{0\} \cup \{1\}$ would also be countable.)
A: Your second argument is closer to being right. The set of all rational numbers cannot include the irrational numbers, no matter how close the rations can get to an irrational. Of course, this doesn't prove that $(0,1)$ is uncountable, but the Cantor diagonalization does.
A: Well, although no-one wants to advocate it, there's always knowledge by accepting authority.  All textbooks and authorities tell us [0,1] is uncountable.
I'm not saying we should bow to authority but it will give us a guideline where we really ought to be going and if we don't get there what we should look for to see where we went wrong.
So according to authority: "[0, 1] is uncountable because there is no 1-1 correspondence between the natural numbers and [0,1]."  Okay.... let's look at our arguments and see how they compare.
Our argument that [0, 1]: we have a list of fractions that seem to be in 1-1 correspondence to the natural numbers.  Are they?  Yes, they are.  These fractions get infinitely close to every real.  Some will have an infinite sequence that goes toward the reals.  So does this make them countable.
Well, to be countable we must associate each of these numbers to a specific order.  Ex: "pi is precisely the 318,295,143rd such fraction."  We never defined these.  We got infinitely close to pi but none of the values ever actually were pi.
So we failed to show countability. 
The argument for uncountability.  You do argue what was wrong with the above argument, that as the sequence we created only includes fractions it will never include any irrationals.  And that the irrationals will be within an epsilon and epsilon may be infintisimally smalls hints at the idea of uncountability.  But "infintesimally small" can be either countably small or uncountably small.  There are uncountably many epsilons but only countably many 1/n s.
We have simply argued that we haven't succeeded.
So how do we find this 1-1 correspondence or show one can not exist?  We use what is called Cantor's diagonalization.  Look it up.  That should be the final word.
