Clarifying Polygonal Paths in Complex Analysis Introduction to Complex Analysis by H. A. Priestley (Revised ed.) shows a diagram (page 40) which is a polygonal path in a set $G$ (link below). Their proof claims we can polygonally pave our way from $z$ to $a$ by using the fact that open discs centred at $z$ enclose a point $w$ such that the line segment $[z,w]\subset D(z;r),r>0$. I am confident in that notion, but maybe it's their diagram which confuses me.

It displays large line segments paving through the set $G$ which if you were to draw open discs centred from the original point, the open disc is then not contained by $G$. I assume they could have used smaller open discs, therefore, more $w$s to get to $a$.
Any clarification on this topic would help me out a lot. Thank you very much.

I assume the statement you're after really is that $G$ open and connected implies $G$ path-connected which, in turn, implies points in $G$ can, in fact, be connected by polygonal paths.
The first implication is simply the fact that $G$ is locally path-connected (connected and locally path-connected spaces are, necessarily, path-connected).
To get the second implication, pick some path $\gamma$ from $z$ to $a$. Since $\gamma$ is compact and $\partial G$ is closed, there exists some $\varepsilon>0$ such that $d(\gamma,\partial G)>\varepsilon$. Accordingly, $(D(x,\varepsilon))_{x\in \gamma}$ is an open cover of $\gamma$ in $G$ and we can apply compactness to get that $z,x_1,...,x_n,a$ such that $(D(x_j,\varepsilon))_{1\leq j \leq n}$ forms a cover of $\gamma$ in $G$. We may, without loss of generality, assume that the ordering of the $x_j$ is such that $D(x_j,\varepsilon)\cap D(x_{j+1},\varepsilon)\neq \emptyset$ for every $j$, $z\in D(x_1,\varepsilon)$ and such that $a\in D(x_n,\varepsilon)$. Note that if our initial path was poorly chosen, this might imply that the $x_j$'s are non-distinct but that doesn't matter for the argument.
At this point, however, it's clear that there is a straight line from $z$ to $x_1$ in $G$, a straight line of $a$ to $x_n$ in $G$ and that there is a polygonal path in $G$ consisting of two line segments between $x_j$ and $x_{j+1}$ for every $j$. Concatenating these paths yields a polygonal path between $z$ and $a$.
