Does the limit of a descending sequence of connected sets still connected? Given a descending sequence of sets 
$$
F_1\supset F_2\supset\cdots F_n\supset\cdots 
$$ 
in which each $F_i$ is connected. I wonder if the limit set
$$
F=\bigcap_{i=1}^\infty F_i
$$
is still connected? I believe it is, but cannot make a proof. Anyone can help?

Updated:
Samuel has showed a counter example.
Thus now I wonder can I add some constraints such that the conclusion holds?
I ask this problem because when I look up the The Princeton Companion to Mathematics，chapter IV.14. Dynamics, section 2.8 The Mandelbrot Set M, there is the following words:

It follows from the above that as $t$ approaches zero, the equipotential
  of potential $t$, together with its interior, gets closer and closer
  to M: that is, M is the intersection of all such sets. Hence, M is a
  connected, closed, bounded subset of the plane.

I wonder why such argument shows $M$, the Mandelbrot set, is connected.

No. Let $F_n$ be the the plane $\mathbb R^2$ minus the line $\{0\}\times(-\infty,n)$.
Added: It is true when all the $F_n$ are compact subsets of $\mathbb R^N$. Suppose otherwise: then there exist open disjoint sets $A,B$ such that $F$ contains points of both $A$ and $B$ and $F$ is contained in $A\cup B$. Now consider $F_n\cap (\partial A)$. Since each $F_n$ is connected, and contains points in both $A$ and $B$, the intersection $F_n\cap (\partial A)$ must be nonempty, and moreover, for $n=1,2,3,\ldots$ it is a decreasing sequence of compact sets, and therefore the intersection of all $F_n\cap (\partial A)$ is nonempty. Contradiction. Thus $F$ is connected.
