Are intersections of simple subsets of $\mathbb{R}^2$ necessarily simple? Edit: Consider the notion of "simply connected," and drop the "connected". Then we obtain the following concept.
Call a subset $A$ of a topological space simple if for all $x,y \in A$ such that there exists a path between $x$ and $y$ lying completely in $A$, it holds that any two such paths are equivalent up to continuous deformation.
Is it always the case that the arbitrary intersection of simple subsets of $\mathbb{R}^2$ is itself simple? Note that this fails if $\mathbb{R}^2$ is replaced by the sphere.
If so, define that the simplification of any $A \subseteq \mathbb{R}^2$ is the intersection of all simple supersets of $A$. If $A$ is connected, is its simplification necessarily connected?
 A: The result becomes easy by using a nice (not fully trivial) result:

A set $A\subseteq \mathbb R^2$ is simple if and only if its complement in $\overline{\mathbb R^2}:=\mathbb R^2\cup\{\infty\}$ is connected.

Now if $A_i$ is simple for each $i\in I$, then 
$$ \overline{\mathbb R^2}\setminus \bigcap _{i\in I} A_i = \bigcup_{i\in I}\left(\overline{\mathbb R^2}\setminus A_i\right)$$
is the union of connected subsets that all have the point $\infty$ in common and hence is connected.
A: An arbitrary intersection of simple sets is simple: if $\gamma, \gamma'$ are two coterminal paths in the intersection, then they belong to each of the original sets, and so the interior of the set bounded by these paths belongs to each original set; this holds (and is well-defined to begin with) because we're dealing with $\mathbb{R}^2$. Therefore the interior belongs to the intersection, and hence the two paths can be continuously deformed to one another within the intersection.
The fact that the simplification of a connected $A$ is connected is trivial.  Suppose $S$ is its simplification. Let $S'$ be the connected component of $S$ which contains $A$; there must be a single one because $A$ is connected.  Then $S'$ is a union of path-components of $S$ (since path-connected implies connected), and thus $S'$ must be simple as well, so $S \subseteq S'$.  Well then $S = S'$, so the simplification of $A$ is connected.
