Show that $|\mathbb{R}|$=$|[0,1]|$.
If we were to find a function whose domain is $\mathbb{R}$ and range is $[0,1]$ and show it is a bijection, then we can show that this is true. The function that I came up with is $f(x)=\frac{\arctan x+\frac{\pi}{2}}{\pi}$, but this function's range is $(0,1)$. Is there a function that would work?
The easiest way is to use what you have and find a bijection between $(0,1)$ and $[0,1]$, then compose the two bijections to get what you need. That question has been asked before. A simple one $[0,1]$ to $(0,1]$ is $$f(x)=\begin {cases} \frac 15&x=0\\ \frac 1{5^{n+1}}&x=\frac 1{5^n}\\x&\text{otherwise} \end {cases}$$ Basically you just push the extra point down the line. You can do the same from the other end to take care of $1$. You get used to ignoring stray points when you have a bijection between infinite sets, as you can always absorb them.
Having $|(0,1)|=|\mathbb R|$ as showed by you is already sufficient, because notice that $$(0,1) \subseteq [0,1] \subseteq \mathbb R$$ Thus there exist injection mapping $f(x)=x$ for those sets, which means: $$|(0,1)| \le |[0,1]| \le |\mathbb R|$$ So it has to be the case that $$|(0,1)| = |[0,1]|=|\mathbb R|$$
This is thanks to Cantor-Bernstein Theorem
Send $0 \to 0$ and consider $f(x) = \frac{1}{x}$. Then $(0,1) \to (1, \infty)$. Try to construct $g$ based on $f$ that satisfies your requirements. You will need translations and scaling.
