Proving that all entire & injective functions take the form $f = ax + b$?

The first part of what you say its right, thats what the author is trying to do.

For the other 2 specific questions.

1.$f$ Is bounded because the closed circle is a compact set and $f$ is continuous.

2.Because the set $f(\{|z|<r\})$ is open it means that there is some ball inside $f(\{|z|<r\})$ that only take points from $f(\{|z|<r\})$, now because Cassorati-Weierstrass assures you that $f(\{|z| > r \}$ is dense in all $\mathbb{C}$, it must have some point inside that ball, thus the intersection, $f(\{|z| > r \} \cap f(\{|z|<r\})$ is non-empty.

1. If $0$ is a removable singularity of $g$, Riemman's theorem shows that there is an open disk $D:=D(0,\frac{1}{r})$ in which $g$ is bounded, for some $r>0$. Note that $$ \textstyle \{|z| < \frac{1}{r}\} = \{\frac{1}{|z|} > r\}, $$ and so $g(D)$ is bounded if and only if $f(\mathbb{C} - D)$ is bounded. But $f$ is also bounded on $D[0,r]$, because this is a compact set and we may use the extreme value theorem (any continuous function $K\to\mathbb R$ from a compact set attains maxima and minima).

2. Denseness is equivalent to intersecting any non-empty set. The open mapping theorem assures that $f(\{ |z| < r)$ is open, and therefore intersects any dense set.