Let $M(I,X)$ denote continuous maps from $I\to X$ where $I=[0,1]$.
My text wants me to show that $x\sim y$ iff there exists a $\alpha\in M(I,X)$ where $\alpha(0)=x$ and $\alpha(1)=y$.
This sounds ultra reasonable. But then when proving reflexivity we let $\alpha(t)=x$ for all $t\in I$. Nothing is ever said about $X$ being a $T_1$ space? So this constant map isn't even necessarily continuous? Since $\alpha^{-1}(x)=I$ which is closed, but perhaps $\{x\}$ is open and not closed?
I guess I am also implicitly assuming they are putting subspace topology from Euclidean on $I$?
Let $f:X\to Y$ be a constant map $f(X)=\{y\}$. Then for any open $U$, either $y\in U$ or $y\not\in U$. If $y\in U$, then $f^{-1}(U)=X$, i.e. the preimage is the entire domain of $f$. If $y\not\in U$, then $f^{-1}(U)=\emptyset$. It holds that $X,\emptyset\in \tau_X$ by the definition of a topology, and hence the constant map is always continuous.