Here is one proof, using the Uniformization Theorem. This proof will be easier in the setting of the "Primer" since the authors are considering universal covering spaces of complete hyperbolic surfaces with geodesic boundary.
Start with a simply-connected surface with boundary $S$ and let $DS$ denote the double of $S$ along its boundary. Then $DS$ admits an involution $\tau$ fixing $\partial S\subset DS$ pointwise. This all can be done smoothly. Now, put a $\tau$-invariant Riemannian metric on $DS$; this defines a conformal structure on $DS$ with respect to which $\tau$ is an antiholomorphic involution. Let $X$ denote the universal covering space of $DS$ with lifted conformal structure. Then $\tau$ lifts to antiholomorphic involutions on $X$. By the Uniformization Theorem, $X$ is conformal to the unit disk (which I will equip with the Poincare metric) or complex plane or $S^2$. I will consider the first case since the proof in the two other cases is similar but simpler. The surface $S$ lifts diffeomorphically to a subsurface $Y\subset X$ (since $S$ is simply connected). Each boundary component $c$ of $Y$ in $X$ is fixed by a lift $\sigma_c$ of $\tau$. Since $\sigma_c$ is an antiholomorphic involution of the unit disk, it is a hyperbolic isometry, hence, its fixed-point set is a geodesic in the hyperbolic plane ${\mathbb H}^2$. Thus, $Y$ is a closed convex subset in ${\mathbb H}^2$. Now, switch to the projective (Klein) model of the hyperbolic plane. Every convex subset of ${\mathbb H}^2$ then becomes a convex subset of the Euclidean plane. It is an elementary exercise to see that each closed convex subset of the hyperbolic plane is homeomorphic to the closed unit disk ${\mathbb D}$ in ${\mathbb R}^2$. Under this homeomorphism $cl_{{\mathbb R^2}}(Y)\to {\mathbb D}$, $Y$ maps to the complement to a closed subset of the boundary of ${\mathbb D}$. Since $Y$ is homeomorphic to your surface $S$, you get the statement you are after.