$F= \bigcap_{i=1}^{\infty} F_i$ isn't necessarily connected where $F_{i+1} \subseteq F_i$ and $F_i \subseteq \mathbb{R}^2$ are closed and connected

Question

In my attempt, I first show that $F$ is closed, this is since we can write $F= \bigcap_{i=1}^{\infty} F_i = (\bigcup_{i=1}^{\infty} F_i^C)^C$ and $\bigcup_{i=1}^{\infty} F_i^C$ is a union of open sets, hence it is open, hence $F=(\bigcup_{i=1}^{\infty} F_i^C)^C$ must be closed.

From here, I am trying to construct an example. I was thinking of $F_i$ being two disjoint disks that are connected with a thick line between them, such that as $i$ grows the line becomes thinner and its thickness tends to zero. The problem is that in this case $F$ still has a one-point width line connecting the disks and this seems not good enough.

Will be happy for any help on this or a better example (Assuming the standart topology on $\mathbb{R}^2$)

Answer 1

Consider taking $F_i$ as the closed upper half-plane with $(0,1)\times [0,i)$ removed.

Answer 2

Another example, $F_i$ us the union of the closed vertical rays starting at $(0,0)$ and $(1,0)$ and the closed half-plane above abscissa $i$. $$ F_i = \{(0,y) : y \geq 0\} \cup \{(1,y) : y \geq 0\} \cup \{(x,y) : y \geq i\} $$ As $i$ increases, the "bridge" between the two rays is yanked away.