Counterexample to Inherited Hausdorff Property of Quotient Space

Lemma. Every singleton set in a Hausdorff space is closed. Proof. For any singleton set $\{p\}$ and an arbitrary point $q$ that is not $p$ , there exist disjoint open sets $U,V$ such that $p\in U,q\in V$ . Vary the point $p$ and take the union of the $V$ s, we have an open set that does not contain the point $p$ and contain any point other than $p$ , the complement of which, the singleton set $\{p\}$ , is closed.

Theorem (Necessary Condition for a Quotient Space to be Hausdorff). If the quotient space $S/\sim$ is Hausdorff, then the elements of any equivalence class form a closed set in $S$ .
Proof. If $\pi:S\to S/\sim$ is the quotient projection map and the quotient itself is Hausdorff, then every singleton set $[p]$ in $S/\sim$ is closed; take the inverse image of $\pi$ and the desired result follows.

Apply the contrapositive of the theorem, we know Option C is definitely a counterexample. Also, more empirically, the points $0$ and $[(0,+\infty)]$ are the points that cannot be separated by disjoint open sets.

That the other two options are Hausdorff can be shown through the definition of Hausdorff space and quotient space.