Suppose $( X , τ_X )$ is path-connected with two distinct points $x \neq y$ . Then is it true that there is a continuous function from $( X , τ_X )$ to $([0,1],τ_)$ with $()=0$ and $()=1$ ?
The definition of path-connectedness gives the existence of a continuous map from $([0,1],τ_E)$ to $(,_)$ satisfying $(0)=$ and $(1)=0$. I do not see any reason for the swapped definition to be true, so I would say it is false and a I would be tempted to find a counterexample with the indiscrete topology on $$ that would be path connected then : by contradiction, if such a map $f$ exists, then $f^{-1}(\{0\})=\emptyset$ so $()\neq0$ or $f^{-1}(\{0\})=X$ so $()\neq1$.
It helps to consider a related notion of arc-connectedness. $X$ is arc-connected iff for each $x \neq y$ in $X$ there is a continuous $f:([0,1],\tau_E) \to X$ such that $f(0)=x, f(y)=1$ and also $f$ is a homeomorphism from $[0,1]$ onto is image $f[[0,1]]$. In that case the answer to your question is yes if we can extend $f^{-1}$ from $f[[0,1]]$ to $X$ (if the image is closed and $X$ is normal Tietze would apply...)
It is well-known that a $T_2$ (Hausdorff) path-connected $X$ is arc-connected and so the answer to your question is yes if $X$ is normal and Hausdorff (so true for all metric spaces e.g.) but not true in general as indeed witnessed by the indiscrete topology and other such trivialities.