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.

H. A. Priestley, Introduction to Complex Analysis, Revised Edition (Oxford University Press 1990), p.39f.:

Let $G_1 = \{z \in G \colon\ \text{there exists a polygonal path in } G \text{ with endpoints } a \text{ and } z\}$ and let $G_2 = G \setminus G_1.$ We require $G_1 = G.$ We shall prove that each of $G_1$ and $G_2$ is open. Connectedness of $G$ will then imply that one of these sets is empty. This cannot be $G_1,$ since $a \in G_1.$
We now establish our claim that $G_1$ and $G_2$ are open. For any $z \in G,$ we can find $r$ such that $D(z; r) \subseteq G.$ For each $w \in D(z; r),$ $[z, w] \subseteq G.$ It follows that $z$ can be joined to $a$ by a polygonal path in $G$ if and only if $w$ can be (see Fig. 3.3). Hence, for $k = 1, 2,$ $z \in G_k$ implies $D(z;r) \subseteq G_k.$

The diagram illustrates the implication that if $z \in G_1$ then $w \in G_1.$ A similar diagram could be drawn to illustrate the converse implication that if $w \in G_1$ then $z \in G_1;$ but that would be overkill. Instead, just imagine that the penultimate segment of the polygonal path from $a$ terminates at $w$ rather than $z.$
