Homeomorphism of the real line-Topology

I need to show that any open interval is homeomorphic to the real line.

I know that $f(x)=a+e^x$ will work for the mapping $f:R \to (a,\infty)$ and $f(x)=b-e^{-x}$ will work for the mapping $f:R \to (-\infty,b).$

Without using two functions, how can I prove the result in general?

Let $$ f(x) = \frac{x}{x^2-1}. $$ This is a homeomorphism from $(-1,1)$ to $\mathbb R$.

Let $$ g(x) = \frac{1}{1+2^{-x}}. $$ That is a homeopmorphism from $\mathbb R$ to $(0,1)$.

For any other bounded interval $(a,b)$, just rescale and relocate.

You've already been given two possible homeomorphisms, but how about another one?

Say that you have two maps $\varphi : A \to B$ and $\psi : B \to C$, both of which are homeomorphisms. It's clear that $\psi \circ \varphi : A \to C$ is again a homeomorphism.

Using this fact, choose your favorite finite open interval $(a,b)$, and prove it is homeomorphic to $\mathbb{R}$.

Next up, take an arbitrary open interval $(c,d)$, and construct a homeomorphism between this an $(a,b)$, and voila, you are done.

In particular, look at the interval $(0,1)$, and its image under the function $\tan(\pi(x-\frac{1}{2}))$. This is pretty clearly a homeomorphism. Now just map an open interval to $(0,1)$ (homeomorphically), and call it a day.

Consider the function $\log(b-x)-\log(x-a)$