Homeomorphism from $(0,1)$ to $\mathbb{R}$

$\tan:(-\frac{\pi}{2},\frac{\pi}{2})\to\mathbb{R}$ is a homeomorphism between $(-\pi/2,\pi/2)$ and $\mathbb{R}$. Define $f:(0,1)\to(-\pi/2,\pi/2)$ by $f(t)=-(1-t)\frac{\pi}{2}+t\frac{\pi}{2}=-\frac{\pi}{2}+t\pi$. Then, $f$ is a homemorphism between $(0,1) $ and $(-\pi/2,\pi/2)$. Therefore, $h:(0,1)\to\mathbb{R}$ given by $h(x)=\tan(f(x))=\tan(-\frac{\pi}{2}+t\pi)$ works.

Would an answer not involving the tangent function fit your needs? If so, try this:

$f:\mathbb{R}\to (-1,1),\quad x\longmapsto \dfrac{x}{1+\mid x\,\,\mid}\,\,$. It's a homeomorphism (check!).

$g:(-1,1)\to (0,1),\quad x\longmapsto\dfrac{x+1}{2}\,\,.$ It's a homeomorphism (check this too!).

Now, set $h:\mathbb{R}\to (0,1)$ by the equation $h(x)=g(f(x))$ for all $x\in\mathbb{R}$. It's a homeomorphism as a compose of two such functions.

The function $$\frac x{\sqrt{1+x^2}}$$ should do nicely.