In my document, I now use the following notations and conventions.
I use $[1,w]$ for real numbered intervals and $\{1,2,...,w\}$ for integer intervals.
If I reuse a specific interval several times, I introduce and name it once, e.g., as $\mathbb{N}_w=\{1,2,...,w\}$, and then refer to it using its name, e.g., $\mathbb{N}_w$.
Thereby, throughout the document, $\mathbb{N}_n$ always denotes an integer interval from $1$ to $n$, while $\mathbb{R}_n$ always denotes a real numbered interval between and including $0$ and $n$.
However, I do not use this notation unless having introduced each specific $\mathbb{N}_?$ and $\mathbb{R}_?$ explicitly.
Using short names for my common intervals, I can now easily and consistently denote multidimensional ranges, e.g., $\mathbb{R}_w \times \mathbb{R}_h$, or express membership, e.g., $x \in \mathbb{N}_w, y \in \mathbb{N}_h$.
I believe that using a subscript index for $\mathbb{R}$ and $\mathbb{N}$ is safe enough and creates the least confusion with other common notations. I now prefer the longer notation $\{1,2,...,n\}$ over $[1..n]$ as it is more common, and over $\mathbb{N}\cap[1,n]$ as it does not require the reader to process the expression, i.e., as it is more explicit.
Yes, your answer is fundamentally wrong. Let me point at that it is not even right in the finite case. In particular, you are using the following false axiom:
If two sets of outcomes are equally large, they are equally probable.
However, this is wrong even if we have just two events. For a somewhat real life example, consider some random variable $X$ which is $1$ if I will get married exactly a year from today and which is $0$ otherwise. Now, clearly the sets $\{1\}$ and $\{0\}$ are equally large, each having one element. However, $0$ is far more likely than $1$, although they are both possible outcomes.
The point here is probability is not defined from cardinality. It is, in fact, a separate definition. The mathematical definition for probability goes something like this:
To discuss probability, we start with a set of possible outcomes. Then, we give a function $\mu$ which takes in a subset of the outcomes and tells us how likely they are.
One puts various conditions on $\mu$ to make sure it makes sense, but nowhere do we link it to cardinality. As an example, in the previous example with outcomes $0$ and $1$ which are not equally likely, one might have $\mu$ defined something like:
$$\mu(\{\})=0$$
$$\mu(\{0\})=\frac{9999}{10000}$$
$$\mu(\{1\})=\frac{1}{10000}$$
$$\mu(\{0,1\})=1$$
which has nothing to do with the portion of the set of outcomes, which would be represented by the function $\mu'(S)=\frac{|S|}2$.
In general, your discussion of cardinality is correct, but it is irrelevant. Moreover, the conclusions you draw are inconsistent. The sets $(0,1)$ and $(0,\frac{1}2]$ and $(\frac{1}2,1)$ are pairwise equally large, so your reasoning says they are equally probable. However, the number was defined to be in $(0,1)$ so we're saying all the probabilities are $1$ - so we're saying that we're certain that the result will be in two disjoint intervals. This never happens, yet your method predicts that it always happens.
On a somewhat different note, but related in the big picture, you talk about "uncountably infinite sets" having the property that any non-trivial interval is also uncountable. This is true of $\mathbb R$, but not all uncountable subsets - like $(-\infty,-1]\cup \{0\} \cup [1,\infty)$ has that the interval $(-1,1)=\{0\}$ which is not uncountably infinite. Worse, not all uncountable sets have an intrinsic notion of ordering - how, for instance, do you order the set of subsets of natural numbers? The problem is not that there's no answer, but that there are many conflicting answers to that.
I think, maybe, the big thing to think about here is that sets really don't have a lot of structure. Mathematicians add more structure to sets, like probability measures $\mu$ or orders, and these fundamentally change their nature. Though bare sets have counterintuitive results with sets containing equally large copies of themselves, these don't necessarily translate when more structure is added.
Best Answer
The renown Cantor's Diagonal Argument shows that although we can easily map the set of natural numbers to the set of real numbers in many ingenious ways, we can never find a surjection $f:\mathbb N\rightarrow\mathbb R$. A surjection here means $f(\mathbb N)=\mathbb R$. Instead, as Cantor proved, we can do no better than $f(\mathbb N)\subset\mathbb R$.
This means that any counting function $f:\mathbb N\rightarrow\mathbb R$ assigning enumerations to the real numbers such that the real numbers can be written as a list $\mathbb R=\{f(1),f(2),...\}$ is impossible. The set $\{f(1),f(2),...\}$ can contain some of the real numbers, but not all.
In this sense, which is called cardinality (see link in comments), the set of real numbers is bigger.
There are various proofs of Cantor's Diagonal Argument. One is to prove the more general statement that given a set $A$, the so-called Power Set $\mathcal P(A)$ containing all subsets of $A$ always has bigger cardinality than $A$.
So assume we have found an enumeration $f:\mathbb N\rightarrow\mathcal P(\mathbb N)$. Then $f(n)$ is some subset of $\mathbb N$ as for instance $\{2,4,6,8,...\}$. If $f$ were surjective it should be possible for any element $B\in\mathcal P(\mathbb N)$ to find some $b\in \mathbb N$ with $f(b)=B$.
Consider $B\in\mathcal P(\mathbb N)$ defined as $$ B=\{n\in\mathbb N\mid n\notin f(n)\} $$ Assume we have found $b$ such that $f(b)=B$. Suppose $b\in B=f(b)$, but then the above definition implies that $b$ should not be an element of $B$. So we must have $b\notin B=f(b)$, but then the above definition tells us that $b$ should be in $B$. Contradiction! For any function $f$ the set $B$ above is well defined, but as just shown no element of $\mathbb N$ can map to it. We must have $f(\mathbb N)\subset\mathcal P(\mathbb N)\setminus B$, so in particular $f(\mathbb N)\neq\mathcal P(\mathbb N)$ for any choice of $f$.
Let us connect the above argument to the real numbers. We can make a simple mapping $g:\mathcal P(\mathbb N)\rightarrow\mathbb R$ that takes a given subset of $B=\{n_1,n_2,...\}\in\mathcal P(\mathbb N)$ of natural numbers and maps it to $$ g(B)=10^{-n_1}+10^{-n_2}+...=\sum_{n\in B}10^{-n} $$ For instance $B=\{2,4,6,8,...\}$ would map to $g(B)=0.010101\overline{01}$. This mapping is injective meaning that different sets $B$ map to different real numbers $g(b)$. So $\mathcal P(\mathbb N)$ has the same cardinality as the subset $g(\mathcal P(\mathbb N))$ of the real numbers meaning that we have the cardinality inequalities $$ |\mathbb N|<|\mathcal P(\mathbb N)|=|g(\mathcal P(\mathbb N))|\leq|\mathbb R| $$ In fact a different choice of $g$ shows that $|\mathcal P(\mathbb N)|=|\mathbb R|$, but that is a different story.