= Solution
A covering-surface formulation of the <Ahlfors second fundamental theorem> is as follows. For a regular finite bordered <Riemann surface> $X$ mapped holomorphically to a fixed metric bordered <Riemann surface> $Y$, let $S=A_f(X)/A(Y)$ be the <average sheet number> and $L$ the pulled-back length of the relative boundary, namely the boundary mapped into the interior of $Y$. With the modern sign convention for the <Euler characteristic>,
$$
\max\{-\chi(X),0\}\ge-S\chi(Y)-hL,
$$
where $h$ depends on $Y$ and its metric, not on $X$ or the map. The associated area comparison is $|S_D-S|\le h_DL$ for each fixed regular target region $D$. These formulations and the sign convention are explained in https://www.math.purdue.edu/~eremenko/dvi/ahl.pdf .
The island version of the <Ahlfors second fundamental theorem> for a disk source, which is the useful form here, says that for $q\ge3$ fixed <Jordan domains> $D_j$ with disjoint closures on the <Riemann sphere>,
$$
\boxed{\sum_{j=1}^q n_j(R)\ge(q-2)S(R)-hL(R).}
$$
Here $n_j$ counts simply connected <islands of a meromorphic function> over $D_j$, once each without weighting by mapping degree, and
$$
S(R)=\frac1{4\pi}\int_{|z|<R}(f^{\#}_{\mathrm{round}})^2dA,\qquad L(R)=\int_{|z|=R}f^{\#}_{\mathrm{round}}|dz|.
$$
An <island of a meromorphic function> is a relatively compact inverse-image component mapped properly onto its target domain; components reaching the source boundary are not counted. The simply connected island estimate is recorded, for spherical disks, in Lemma 2 of https://pdfs.semanticscholar.org/d247/562607e237af7c8fc6e683b77be6b94829fa.pdf . Regular target regions give the same estimate with a region-dependent constant. General <Jordan domains> can be enclosed in slightly larger regular <Jordan domains> still having disjoint closures. A <simple island> over an enlarged domain restricts to one over the original domain, so it suffices to prove the conclusion for regular domains.
Suppose a nonconstant <meromorphic function> on the plane has no <simple islands> over five such domains. On every counted simply connected island its <proper map> has integer degree at least two. The <area formula> therefore gives
$$
2n_j(R)\le S_{D_j}(R),\qquad S_{D_j}(R)=\frac1{A(D_j)}\int_{\{|z|<R\}\cap f^{-1}D_j}(f^{\#}_{\mathrm{round}})^2dA.
$$
Together with area comparison and the <Ahlfors second fundamental theorem>, this yields
$$
3S(R)\le\sum_jn_j(R)+hL(R)\le\frac52S(R)+CL(R),\qquad \frac12S(R)\le CL(R).
$$
The area comparison can also be seen directly. The two-form $[\mathbf1_D/A(D)-1/(4\pi)]dA$ on the target sphere has zero integral and a bounded one-form primitive. One construction solves its <Poisson equation>: the gradient of the spherical <Green function> has an integrable $1/\mathrm{distance}$ singularity, so convolution against the bounded density gives a bounded primitive. Pulling back and applying <Stokes theorem>, or smooth approximation at the domain boundary, bounds the integral by a constant times $L(R)$.
We now derive the needed <length-area exhaustion of the complex plane>. The <Cauchy-Schwarz inequality> on a circle gives
$$
L(R)^2\le2\pi R\int_{|z|=R}(f^{\#}_{\mathrm{round}})^2|dz|=8\pi^2R S'(R).
$$
If $L(R)/S(R)\ge\varepsilon>0$ for all sufficiently large $R$, then $S(R)>0$ and
$$
\left(\frac1{S(R)}\right)'=-\frac{S'(R)}{S(R)^2}\le-\frac{\varepsilon^2}{8\pi^2R}.
$$
Integration makes $1/S(R)$ negative, which is impossible. Hence there are radii $R_k\to\infty$ with $L(R_k)/S(R_k)\to0$. This contradicts $S/2\le CL$ and proves the <Ahlfors five islands theorem>: \b[one of the five target domains has a bounded simply connected inverse-image component mapped biholomorphically onto it].
Four domains do not suffice. Take the <Weierstrass elliptic function> $\wp$ for the square lattice $\mathbb Z+i\mathbb Z$. The equation
$$
(\wp')^2=4(\wp-e_1)(\wp-e_2)(\wp-e_3)
$$
shows the three distinct finite branch values $e_1,e_2,e_3$; the fourth is infinity, because all <poles> are double. Equivalently, the degree-two map from the elliptic torus to the <Riemann sphere> has these four branch values, with all their preimages of local degree two. Choose four small disjoint <Jordan domains>, each containing one of these <four totally ramified Weierstrass values>. A degree-one <island of a meromorphic function> over any one would contain a preimage of its central value of local degree one, contradicting total ramification. Thus \b[$\wp$ provides four domains with no simple island].
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