Disconnected projection components admit a separating Jordan curve

Choose one connected component $P_0$ of $P$. Around each of its finitely many vertices take a small closed disk, and around each edge take a narrow closed strip joining the incident vertex disks; for a crossingless circle take an annular neighborhood. Round the finitely many corners. Because the graph is finite and its distinct components are disjoint compact sets, the disks and strips can be chosen so that their union $N$ is a compact connected subsurface of $S^2$ with smooth finite boundary, contains $P_0$ in its interior, and misses every other component of $P$.

Choose points $p_0\in P_0$ and $p_1$ in another component of $P$. Choose a piecewise-smooth path from $p_0$ to $p_1$ transverse to the finite smooth boundary $\mathrm{\partial }N$. Its endpoints lie respectively in the interior of $N$ and outside $N$, so it crosses $\mathrm{\partial }N$ an odd number of times. The boundary has finitely many disjoint simple closed components; at least one, call it $\gamma$, is met an odd number of times. The curve $\gamma$ lies in $S^2\setminus P$, since $N$ contains $P_0$ in its interior and misses every other component. By [F2], every transverse crossing of $\gamma$ switches sides, so $p_0$ and $p_1$ lie on opposite sides of $\gamma$.

The construction uses finitely many neighborhoods and applies also when $P_0$ is a crossingless circle; therefore it proves the asserted separation for every nonempty disconnected regular projection. Its chosen separating curve is a smooth boundary component of $N$.