Let $X$ be a connected no singleton set in $(Y,\tau_{cof})$ with $Y$ infinite. Show that $X$ is totally pathwise disconnected. (By definition, $X$ is totally pathwise disconnected if the only continuous function $f:[0,1]\rightarrow X$ are the constant maps.)
$X$ is infinite, because the connected sets in the cofinite topology are the singletons and the infinite sets when $Y$ is infinite.
I've been trying to prove this exercise. But i think that a counterexample could be the following function:
$\phi:[0,1]\rightarrow [0,1]$
$\phi(x)=x$
And $\phi$ is continuous because $\tau_{cof}\subseteq\tau_u$ where $\tau_u$ is the usual topology in $\mathbb{R}$.