Here is my proof for the statement that the set of all closed bounded intervals has cardinality $c.$
Let $S$ denote the set of all closed bounded intervals.
We know that any element in $S$ is of the form $[a,b]$ where $a,b \in \Bbb R.$ Define the function $f$ from the set $S$ to $\Bbb R \times \Bbb R$ by sending $[a,b]$ to $(a,b).$
This is obviously a bijection. Since $\Bbb R \times \Bbb R$ has cardinality $c$, we are done.
Can someone have a look at my proof? I am concerning that $f$ may not be a function. Is $f$ really a function?
Well almost.
What you need for an interval $[a,b]$ is not only that $a,b\in \mathbb R$, but also that $a\le b$, that is $S=\{[a,b]: a\le b\}$. Your solution has issues because of that. However that can be fixed.
If we can use Cantor-Bernstein (that is $|A|\le |B|$ and $|B|\le |A$ implies that $|A|=|B|$) we could fix this note that the finite intervals have an bijection to a subset of $\mathbb R^2$ (or that it is a injection), so we know that $|S||\le |\mathbb R^2|=|\mathbb R|$. We also have an surjection (fx $f([a,b])=a$) to $\mathbb R$ so we know that $|S|\ge|\mathbb R|$. Now Cantor-Bernstein shows that $|S|=|\mathbb R|$
If you can't use Cantor-Bernstein you would map the subset of $\mathbb R^2$ to $\mathbb R^2$ bijectively. What we basically need is a bijection $\phi$ that maps $\mathbb R_{\ge0}$ to $\mathbb R$. With that we can form the bijection by mapping $[a,b]$ to $(a, \phi(b-a))$. The only thing is to construct such a bijection $\phi$. This can be done by noting that we can uniquely decompose any number $x = n(x)+\xi(x)$ where $n(x)$ is an integer (which is non-negative if $x$ is) and $0\le \xi(x)<1$. So we define $\phi$ as
$$\phi(x) = \begin{cases} n(x)/2 + \xi(x) & \text{ if } n(x) \text { is even}\\ -(n(x)+1)/2 +\xi(x) & \text{ if } n(x) \text { is odd} \end{cases}$$
Basically we map the integer part using a bijection from $\mathbb N$ to $\mathbb Z$.