It is a common saying that there are more real numbers than integers, since you can not uniquely map integers to real numbers - even after using up all integers, there will still be at least one real number left that you haven't mapped to any integer.
So I applied the same argument to the two intervals in question. One obvious way to map number from $[0,1]$ to $[0,2]$ is $\times 2$ - $0$ is still $0$, $0.25$ becomes $0.5$, and $1$ becomes $2$. Good enough.
However, if we change the second interval to an open interval, namely $(0,2)$, then the number $0$ and $1$ in the original mapping is now the "remaining" number, while all numbers in $(0,2)$ have already been "used". Does that mean that $[0,1]$ have more real numbers than $(0,2)$? (even though $(0,2)$ seems to be bigger than $[0,1]$…)
Does my argument even make sense? I did some Googling but didn't found anything useful.
The map $x \mapsto \dfrac{x+1}{2}$ is a bijection $[0,1] \to [\dfrac12, 1] \subseteq (0,2)$.
Therefore, $(0,2)$ has at least as many elements as $[0,1]$.
The map $x \mapsto \dfrac{x}{2}$ is a bijection $(0,2) \to (0,1) \subseteq [0,1]$.
Therefore, $[0,1]$ has at least as many elements as $(0,2)$.
Thus*, the two intervals have the same number of elements.
(*) To make this rigorous, you can invoke the Cantor–Schröder–Bernstein theorem.