How come this is surjective?

Question

Reference: Hocking & Young - Topology p.55

Let $(X,\tau)$ be a compact connected metric space.

Let $\leq_X$ be a total ordering on $X$ which induces $\tau$ as the order topology.

Let $D$ be a countable dense subset of $X$ and assume that there exists an order isomorphism $f:D\rightarrow \mathbb{Q}$.

Assume that there exists a function $F:X\rightarrow \overline{\mathbb{R}}$ such that $F=f$ on $D$ and $F(p)=\infty$ and $F(q)=-\infty$ and $F(x)$ is real for all $x\neq p,q$ and $F(x)$ is given by $\sup (f( \{y\in D: y<_X x\}))=\inf (f( \{y\in D: y>_X x\} ) )$.

Why is $F$ is surjective in this case? Of course $F$ is an order homomorphism and injective, but I don't see why this is surjective. Moreover, why is it continuous?

Answer 1

Well if you can show it's continuous, I think the surjectivity follows. The continuous image of a connected set is connected, and you already have that the image contains both the "end points" $\infty$ and $-\infty$.