A maximal subset of $S^2$ with respect to a connectedness property Let the set $A$ be a circle with a chord on the sphere $S^2$. Obviously $A$ has the following property:

P: $\quad$ Any two points $a$ and $b$ of $A$ can be connected by a path that
           meets no other points of $A$ than $a$ and $b$.

Besides $A$ is maximal, no bigger set containing $A$ has P. 
I guess that every maximal set of $S^2$ having P is homeomorphic to $A$ but how to prove that?
 A: As shown in the comment of @Vladimir, the answer is "no" without further hypotheses.
But under a reasonably nice further hypothesis about $A$ the answer will be "yes", namely if $A$ satisfies the following:
(1) $A \subset S^2$ is an embedded finite graph, and for every component $U$ of $S^2-A$ there is a Jordan curve $c \subset A$ such that $U$ is a component of $S^1-c$. 
(By the way this hypothesis rules out the figure 8 of my comment).
Under hypothesis (1) I'll prove that if $A$ satisfies P, and if $A$ is maximal with respect to P, then $A$ is a Jordan curve with a chord attached.
It follows from (1) that every point of $A$ lies on a Jordan curve contained in $A$. It also follows from (1) that $A$ is connected, because otherwise some component of $S^2-A$ has disconnected frontier. And since $A$ is finite graph which is a connected union of topological circles, it follows that any two points of $A$ lie on an embedded circle in $A$, i.e. on a Jordan curve in $A$. 
Pick a Jordan curve $c \subset A$, which exists because of (1). Since $A$ is maximal with respect to P it follows that $A \ne c$. Pick points $q \in c$ and $p \in A-c$, pick a Jordan curve $c' \subset A$ containing $p$ and $q$, and let $a \subset c'$ be the maximal compact arc such that $p \in a$ and $a \cap c = \{x,y\}$ is the set of endpoints of $a$. In other words, $a$ is a "chord of $c$". 
Although it is possible that $x = y$, and so $c \cup a$ is a "figure 8", that possibility can be ruled out using (1) with a separate argument. At the moment let me focus on the case $x \ne y$.
I'll prove that $A = c \cup a$. The points $x,y$ subdivide $c$ into two arcs $c=c_1 \cup c_2$ intersecting only in their common endpoints $x,y$. The set $S^2 - (c \cup a)$ has three components: the component $U_1$ with frontier $c$; the component $U_2$ with frontier $c_1 \cup a$;  and the component $U_3$ with frontier $c_2 \cup a$. Suppose that $A \ne c \cup a$, and pick a point $x \in A - (c \cup a)$. I'll go through the cases $x \in U_1$, $U_2$, or $U_3$ to pick a point $y \in A$ that is separated from $x$ by a Jordan curve contained in $A$, and therefore any path with endpoints $x,y$ crosses $A$ contradicting P. If $x \in U_1$ then $c$ separates $x$ from any $y$ in the interior of the arc $a$. If $x \in U_2$ then $c_1 \cup a$ separates $x$ from any $y$ in the interior of $c_2$. And if $x \in U_3$ then $c_2 \cup a$ separates $x$ from any $y$ in the interior of $a_1$.
In the case $x=y$ where $c \cup a$ is a figure 8, there are still three components of $S^2 - (c \cup a)$: $U_1$ with frontier $c$; $U_2$ with frontier $c \cup a$; and $U_3$ with frontier $a$. By a similar argument to what's above, one can show that $A \ne c \cup a$ leads to a contradiction. But also $A = c \cup a$ contradicts (1) because $c \cup a$ is the frontier of $U_2$ but is not a Jordan curve.

By the way, although hypothesis (1) might seem rather strong, there is a theorem of Carsten Thomassen that says (1) is equivalent to seemingly much weaker assumptions about $A$, namely:
(2) $A$ is a compact subset of $S^2$;
(3) Every point of $A$ is on the frontier of at least two components of $S^2-A$;
(4) For every point $x \in A$ and every component $U$ of $S^2-A$ such that $x$ is on the frontier of $U$, there is a continuous embedding $\gamma : [0,1] \to S^2$ such that $\gamma[0,1) \subset U$ and $\gamma(1) = x$. 
The equivalence (1) $\iff$ (2,3,4) is proven by Thomassen in his paper "The converse of the Jordan curve theorem and a characterization of planar maps". 
