Is $\mathbb R$ endowed with countable complement topology path connected ? I only know that it is connected . Please help . Thanks in advance
No. Note that non-singleton, countable sets in the co-countable topology are disconnected. Indeed, let $E$ be at most countable and non-singleton [and non-empty], and let $x \in E$. Then both $\{x\}$ and $E \setminus \{x\}$ are open, disconnecting $E$.
Now, let $a,b \in \mathbb R$, $a \neq b$. Suppose there exists $f : [0,1] \to \mathbb R$ which is continuous from the Euclidean topology to the co-countable topology and has $f[0]=a$ and $f[1]=b$. Let $D \subset [0,1]$ be a countable, dense subset. Since $f[[0,1]]$ is connected, and is non-singleton since it contains both $a$ and $b$, it is uncountable. Therefore $f[[0,1]] \setminus f[D]$ is non-empty, and since $f[D]$ is at most countable, it is open. Thus $f^{-1}[f[[0,1]] \setminus f[D]] \subset [0,1] \setminus D$ is non-empty and open. But this contradicts the density of $D$.