Counterexample of path-connectedness "swapped" definition

Question

Suppose $(X, \tau_X )$ is path-connected. Then is it true that there is a continuous function from $(X, \tau_X )$ to $([0,1],τ_E)$ with $f(x) = 0$ and $f(y) = 1$ ?

My answer is no, the counterexample I give is $X=[0,1) \cup (1,2]$, $x=0$ and $y=\frac\pi2$, so that $f(z)=\sin(z)$ satisfies $f(0)=0$ and $f(\frac\pi2)=1$ and $f$ is clearly continuous.

I guess there is a much easier counterexample, any ideas ?

Answer 1

Let $X$ be a topological space which consists of a single point. Then $X$ is path-connected, but the range of any map from $X$ to $[0,1]$ consists of a single point.

Answer 2

If $X$ is $\Bbb R$ in the indiscrete (trivial) topology, then $X$ is path-connected (any map into an indiscrete space is continuous) but there is no non-constant continuous map to $([0,1], \tau_E)$.