Prove that $\mathbb{N}$ with cofinite topology is not path-connected space.

U can easily show that:

$[0,1]\subset \mathbb R$ can not be written as countable union of closed and disjoint sets $\implies $ $\mathbb N$ with the cofinite topology is not patch-connected

Prove: If $\mathbb N$ where patch-connected then there is a continuous map $\gamma :[0,1] \rightarrow \mathbb N$ such that $\gamma(0)=x$ and $\gamma(1)=y$ for all x,y $\in \mathbb N$. Then $[0,1]=\gamma^{-1}(\mathbb N)=\gamma^{-1}(\bigcup_{i=1}^{\infty}\{i\})=\bigcup_{i=1}^{\infty}\gamma^{-1}\{i\}$ Since finite elements in the confinite topology are closed and $\gamma$ is continuous we have a countable union of closed disjoints sets. A contradiction.

Now it is enough to prove that there does not exist such an union.

Here is the basic idea: (Again by contradiction)

1) Suppose that $\bigcup_{i=1}^{\infty}B_n=[0,1]$ with closed and disjoint $B_n$

2) Construct a decreasing sequence of closed intervals $...I_4 \subset I_3 \subset I_2 \subset I_1 \subset [0,1]$ such that $B_n \cap I_n =\emptyset$

3) The set $\bigcap_{n=1}^{\infty}I_n$ is not empty. Let $x$ be an element of the set. Then $x$ is element of every $I_n$ and of exactly one $B_n$, hence $B_j \cap I_j=\{x\}$ for one $j\in \mathbb N$. A contradiction. The statement follows.

I left the construction of the intervals $I_n$ by the reader :)