Looking for a counter example for non-connected intersection of descending chain of closed connected sets

Question

Let $X$ be a topological space and let $\left\{ Y_{i}\right\} _{i=1}^{\infty}$ be a descending chain of closed connected subsets of $X$. I know from reading elsewhere that ${\displaystyle \bigcap_{i=1}^{\infty}Y_{i}}$ is not necessarily a connected subspace of $X$ but I have no counter example and I haven't managed to come up with one. There is a counter example here to the same question while also assuming $X$ is compact. However, it uses the quotient topology which I haven't studied about so I would prefer a different counterexample that does not use the quotient topology.

Help would be appreciated!

Answer 1

The answer here can also be said without the quotient:

Let $X:=[-1,1]\cup\{0'\}$ where $0'$ is a new, fictive element, playing the role of a second origin, and define the topology on $[-1,1]$ as usual, and let $(-a,b)\setminus\{0\}\cup\{0'\}$ be a base for open neighborhoods of $0'$.

Now consider $Y_n:=[-1/n,1/n]\cup\{0'\}$.

Answer 2

Try the following sets: $$Y_i = \bigg([-2,2]\times\mathbb{R}\bigg) - \bigg((-1,1)\times(-i,i)\bigg).$$

That is, $Y_i$ is a closed infinite strip with successively larger open boxes removed. The intersection $$\bigcap_{i=1}^\infty Y_i = ([-2,-1]\cup[1,2])\times\mathbb{R}$$ is disconnected, but each $Y_i$ is closed and connected.