In this post: Understanding a proof about nested nonempty connected compact subsets they threat the case when every connected subspace is also closed and $X$ is compact.
Is there any related proof about general non compact and non closed connected subsets?
If not are there any counter examples available? I mean a nested sequence of nonempty connected subset such that the intersection is not connected. Thanks in advance.
Here's another example of a family of nested, connected, non-compact sets that have a disconnected intersection, this time in a metric space.
Consider $\ell^2$, the vector space of real, square-summable sequences, with the usual norm $\|(x_n)\| = \sqrt{\sum x_n^2}$. Let $X$ be the unit sphere of $\ell^2$, specifically the set of points $(x_n)$ where $\|(x_n)\| = 1$. Further, for each $n \ge 1$, let $$X_n = \{(x_m)_{m=1}^\infty \in X : x_2 = x_3 = \ldots = x_{n+1} = 0\}.$$ Then $X_n$ is (path) connected. It's a bit messy coming up with specific paths connecting any two points, but there is a trick you can use.
Suppose $(x_m) \in X_n \setminus \{-(e_m)\}$. Let $(e_m)$ be the point in $\ell^2$ such that $e_1 = 1$ and $e_m = 0$ for $m > 1$. For $K > 0$, consider the point $$(y_m) = \frac{(x_m) + K(e_m)}{\|(x_m) + K(e_m)\|} \in X_n.$$ As $K \to \infty$, $(y_m) \to (e_m)$ continuously, proving that $(e_m)$ is in the same connected component as $(x_m)$.
If $(x_m) = -(e_m)$, then form a continuous path: $$t \mapsto (\cos(t), 0 \ldots, 0, \sin(t), 0 \ldots) \quad t \in [0, \pi],$$ where the $\sin(t)$ term occurs in the $(n+2)$th coordinate, from $(e_m)$ to $-(e_m)$, completing the proof that $X_n$ is connected.
Now, $$\bigcap_{n=1}^\infty X_n = \{(x_m) \in X : x_m = 0 \; \forall m > 1\} = \{(e_m), -(e_m)\},$$ which is disconnected.
An easier counter-example would be taking
$$X_i=\Bbb R^2\setminus\left(R_i\times\{0\}\right)$$
where
$$R_i=\{x\in \Bbb R\mid x<i\} \text{ (the left ray)}$$
Then
$$\bigcap_{i\in \Bbb N} X_i=\text{upper open half plane + lower open half plane}$$
Cleraly it's not connected.
