What is the degree of a homeomorphism from $S^1$ to $S^1$? If $f : S^1 \to S^1$ is a homeomorphism, what is the degree of it?
I was told the answer is 1 or -1, but I can't prove it. Can anyone help me?

It would help if you told us which definition of degree are you using, as Jim Conant asked you.
Anyway, one of the properties of your degree must be being multiplicative with respect to composition. That is:
$$
\mathrm{deg}\ (g\circ f) = \mathrm{deg}\ g \cdot \mathrm{deg}\ f
$$
Moreover:
$$
\mathrm{deg}\ \mathrm{id} = 1 \ .
$$
So, if $f: S^1 \longrightarrow S^1 $ is a homeomorphism, you have an inverse $f^{-1}$. Applying the degree to the equality
$$
f \circ f^{-1} = \mathrm{id}
$$
you get 
$$
\mathrm{deg}\ f \cdot   \mathrm{deg}\ f^{-1}  = 1  \ .
$$
That is, $\mathrm{deg}\ f$ must be an integer number which has an inverse (namely, $\mathrm{deg}\ f^{-1} $). But the only integer numbers with an inverse are $\pm 1$.
