Let all $f:(X,T) \rightarrow (\mathbb{R}, st)$ be continuous. Show $T$ discrete. Let all $f:(X,T) \rightarrow (\mathbb{R}, st)$ be continuous for every $f$. Show $T$ discrete. $st$ is the standard topology. **My logic:** So we know that for all $U \in (\mathbb{R}, st)$, that $f^{-1}(U)$ is open in $(X,T)$ Given any $V \subset X$. Then there exists a continuous function $g$ such that $g^{-1}(W) = V$ for some $W \in (X,T)$ We can do this because of continuity, we can just form a union or intersection of $W$s which generated this $V$ through the preimage of $g$. Very wishy washy I know. But is my intuition correct? Can we make it more rigorous? It was trivial for me to show That given $T$ discrete that all functions are continuous. The other way around isn't so trivial.

I think you're on the right track. To make the argument more precise, for each subset $E$ of $X$ consider the function $f_E$ defined by $f_E(x)=1$ if $x\in E$ and $f_E(x)=0$ otherwise. What is $f_E^{-1}((0,\infty))$?