Complete proof of PAC learning of axis-aligned rectangles
Let $R_S, R_T$ denote the sample rectangle (smallest consistent rectangle) and the target rectangle correspondingly. Since $R_S\subseteq R_T$, $R(R_S)\le \mu\left(R_T\setminus R_S\right)$. The intuition behind taking strips of mass $\epsilon$ around the edges of $R_T$ is that you want to satisfy two conditions:

The strips, $r_1,...,r_4$ should have large enough mass so that with high probability $m(\epsilon)$ samples do not miss them (this would follow from a simple bound for a geometric random variable).

The strips should have small enough mass, so that if all were seen by the samples, then $R_T\setminus R_S\subseteq \bigcup\limits_i r_i$, where the latter union has small probability thus bounding $R(R_S)$.


Your question is settled by the geometric argument in the second point. If all strips were seen by the samples, then we can only err on points in the strips.