Composition of orientation preserving and reversing homeomorphisms Let $S^1$ be the unit circle and $f:S^1 \to S^1$ is a homeomorphism.
We say $f$ is a orientation preserving homeomorphism if any lifting of $f$ to the covering space $\mathbb{R}$ is
strictly increasing and It is called orientation reversing if any lifting of $f$ to the covering space $\mathbb{R}$ is strictly decreasing.
Now I have some question about them :
1- I want to prove that any homeomorphism $f:S^1 \to S^1$ is either orientation preserving or else orientation reversing.
2- The composition of an orientation preserving homeomorphism and an orientation reversing is orientation reversing.
3- If $a,b \in S^1$ then $f$ is orientation preserving if $f(a,b)=(f(a),f(b))$.
And also why the degree of a homeomorphism is either $1$ or $-1$ and does it related to the question 2?
because we know $deg(fog)=deg(f)deg(g)$
 A: Let us first show

Each continuous injection $\phi : J \to \mathbb R$ defined on an interval $J \subset \mathbb R$ is either strictly increasing or strictly decreasing. (Note that intervals may be open, half-open, closed / bounded, unbounded.)

Proof. Let $H = \{(x,y) \in J \times J \mid x < y \}$. It is easy to verify that this is a convex subset of $\mathbb R^2$, thus it is path connected and a fortiori connected.
Let $A = \{(x,y) \in H \mid \phi(x) < \phi(y) \}$ and  $B= \{(x,y) \in H \mid \phi(x) > \phi(y) \}$. Clearly $A \cap B = \emptyset$ and $A \cup B  = H$ (note $\phi(x) = \phi(y)$ implies $x = y$ because $\phi$ is injective). Since $\phi$ is continuous, both $A$ and $B$ are open in $J \times J$, thus also open in $H$. Since $H$ is connected, one of $A$ or $B$ must be $= H$ and the other $= \emptyset$. This means
that $\phi$ is either strictly increasing or strictly decreasing.
Next let us show

If $H : \mathbb R \to  \mathbb R$ is a lift of a homeomorphism $h : S^1 \to S^1$, then $H$ is a homeomorphism.

Proof. Let $p : \mathbb R \to S^1, p(t) = e^{2\pi it}$, be the standard covering map. As a lift of $f : S^1 \to S^1$ we denote any map $F : \mathbb R \to \mathbb R$ such that $p \circ F = f \circ p$.
You certainly know that if $F,F'$ are lifts of $f$, then
$$(*) \phantom x F'(t) = F(t) + k \text{ for all } t \text{ with a } \textbf{fixed } k \in \mathbb Z. $$
In fact, we have $e^{2\pi i(F(t) -F'(t))} = e^{2\pi iF(t)}/e^{2\pi i F'(t)} = (p \circ F)(t)/ (p \circ F')(t) = (f \circ p)(t)/ (f \circ p)(t) = 1$, thus $(F - F')(t) = F(t) -F'(t) \in \mathbb Z$ and by continuity of $F - F'$ we see that $(F - F')(t) = k$ for some fixed $k \in \mathbb Z$. This means that $F' = \tau_k \circ F$ with the translation homeomorphism $\tau_k : \mathbb R \to \mathbb R, \tau_k(t) = t + k$.
If $F, G$ are lifts of $f, g$, then $p \circ G \circ F = g \circ p \circ  F = g \circ f \circ  p$, thus $G\circ F$ is a lift of $g \circ f$.
Let $h$ be a homeomorphism with inverse homeomorphism $h^{-1}$ and let $H, \bar H$ be lifts of $h, h^{-1}$. Then $\bar H \circ H$ is a lift of $h^{-1} \circ h = id$. Since also $id : \mathbb R \to \mathbb R$ is a lift of $id : S^1 \to S^1$, we get $(\tau_k \circ \bar H ) \circ H = \tau_k \circ (\bar H  \circ H) = id$ for some $k \in \mathbb Z$. Similarly we get $H \circ \bar H = \tau_r \circ id = \tau_r$ for some $r \in \mathbb Z$. The latter implies $H \circ (\bar H \circ \tau_r^{-1}) = id$. Thus $H$ has a left inverse $H' = \tau_k \circ H$ and a right inverse $H'' = \bar H \circ \tau_r^{-1}$. But now $H'' = id \circ H'' = H' \circ H \circ H'' = H' \circ id = H'$, thus $H$ is a homeomorphism with inverse $H^{-1} = H' = H''$.
Your question 1 is answered by the above two theorems.
By the degree formula $\deg(f \circ g) = \deg(g)\deg(f)$ we see that any homeomorphism $h$ has degree $\pm 1$ (since $\deg(id) = 1)$). In fact, $\pm 1$ are the only elements of $\mathbb Z$ which have a multiplicative inverse. Let $H$ be a lift of $h$. It is a homeomorphism, thus $H$ is either strictly increasing or strictly decreasing. In the first case it must have positive degree, in the second case negative degree. Thus

A homeomorphism is orientation preserving iff it has degree 1; it is orientation reversing iff it has degree -1.

Thus the degree formula also answers your question 2.
Concerning 3. : It is not really precise how you define "open intervals" $(a,b) \subset S^1$. It seems that if $a, b \in S^1$ are two distinct points, then you move counterclockwise from $a$ to $b$ and all points strictly between $a$ and $b$ constitute $(a,b)$. I think my answer to Open sets on the unit circle $S^1$ explains it more precisely. The open intervals $(a,b) \subset S^1$ are precisely the images $p((s,t))$ of open interval $(s,t) \subset \mathbb R$ such that $0 < t - s < 1$, where we have $a = p(s)$ and $b = p(t)$. Let us show that an orientation preserving homeomorphism $h$ maps $(a,b)$ onto $(h(a),h(b))$.
Clearly $h(a) \ne h(b)$. Let $H$ be a lift of $h$. Then $H(s) < H(t)$ and $H$ maps $(s,t)$ homeomorphically onto $(H(s),H(t))$. We have $p(H(s)) = h(a), p(H(t)) = h(b)$ and $0 < H(t) - H(s) < 1$. Concerning the last inequality: If $H(t) - H(s) = 1$, then $h(b) = h(a)$ which is impossible. If $H(t) - H(s) > 1$, then $p \mid_{(H(s),H(t))}$ is not injective, thus $p \circ H \mid_{(s,t)}$ is not injective which is a contradiction since $p \circ H \mid_{(s,t)} = h \circ p \mid_{(s,t)} = h \mid_{(a,b)}$.
