Exercise 4, Section 23 of Munkres’ Topology 
Show that if $X$ is an infinite set, it is connected in the finite complement topology.

My attempt:
Approach(1): Assume towards contradiction, $X$ is not connected. Then $\exists f:X \to \{0,1\}$ such that $f$ is continuous and $f$ is not constant. So $f^{-1}(\{0\})\neq \emptyset$ and $f^{-1}(\{1\})\neq \emptyset$. It is easy to check $X-f^{-1}(\{0\})= f^{-1}(\{1\})$ and $X-f^{-1}(\{1\})= f^{-1}(\{0\})$. In fact $f^{-1}(\{0\})$ and $f^{-1}(\{1\})$ form a separation of $X$. $\mathcal{T}_{X}=\{ U \subseteq X|$ either $X-U$ is finite or $X\}$. Since $f$ is continuous, $f^{-1}(\{0\}), f^{-1}(\{1\})\in \mathcal{T}_X$. So $X-f^{-1}(\{0\})$ and $X-f^{-1}(\{1\})$ is finite or $X$. If $X-f^{-1}(\{0\})=X$, then $f^{-1}(\{0\})= \emptyset$, since $f^{-1}(\{0\})\subseteq X$. Which contradicts our initial assumption of $f^{-1}(\{0\})\neq \emptyset$. Similarly $X-f^{-1}(\{1\})\neq X$. So $X-f^{-1}(\{0\})$ and $X-f^{-1}(\{1\})$ is finite. Which implies $f^{-1}(\{1\})$ and $f^{-1}(\{0\})$ is finite. So $f^{-1}(\{0\}) \cup f^{-1}(\{1\})= f^{-1}(\{0\}\cup \{1\})=f^{-1}(\{0,1\})=X$. So $X$ is finite. Which contradicts hypothesis of problem. Hence $X$ is connected. Is this proof correct?
Approach(2): Show that if $X$ is a infinite set, it is connected in the finite complement topology. Let $A$ and $B$ form a separation of $X$. From separation definition of connectedness, it is easy to check $X-A=B$ and $X-B=A$. Since $A,B\in \mathcal{T}_X$, we have $X-A=B$ and $X-B=A$ is finite or $X$. By definition of connected space, $A,B\neq X$. Now if $A,B$ is finite, then $A\cup B=X$ is finite. Thus we reach contradiction.
 A: If you want to use the function: let $f: X \to \{0,1\}$ be continuous. If $f$ is not constant, the sets $F_0:=f^{-1}[\{0\}]$ and $F_1:=f^{-1}[\{1\}]$ are both closed by continuity, and not $X$ (or $f$ would be constant) so they must be equal to a finite set in $X$ (in the finite complement topology the closed sets are precisely the finite sets and $X$ and that's it), while  $F_0 \cup F_1 = X$. As $X$ is infinite this is impossible. So $f$ is constant and $X$ is connected.
It's the approach 2 way in an alternative form..
A: Approach (2) looks correct to me. Instead of using a "separation", you can also use the existence of non-trivial clopen sets.
Let's call a clopen set trivial if it is $\emptyset$ or $X$.
A closed set different from $X$ is finite.
An open set different from $\emptyset$ is infinite.
A non-trivial clopen set has to be finite and infinite. So, there are none. A topological space where the clopens are only the trivial ones is connected.
But I like very much your conclusion that $X$ has to be finite. :-)

And this is also approach (1)!!! Because, actually, $f^{-1}(1)$ is a clopen set.
So, I would not recommend using $f^{-1}(0)$ or $f^{-1}(1)$, because at the end, you are talking about a clopen set, or about a "separation": $f^{-1}(0)$ and $f^{-1}(1)$. Why use $f^{-1}$, then? :-)
