Hausdorff spaces from continuous functions I am trying to solve Exercise 2.35 in John M. Lee. Introduction to Topological Manifolds, p. 32:

Let $X$ be a topological space. Assume that for every $p\in X$ there exists a continuous function $f:~X\longrightarrow\mathbb{R}$ such that $f^{-1}(0)=\{p\}$. Show that $X$ is Hausdorff.

(The inverse $f^{-1}$ here is implied as converse, not a bijective inverse.)
My thinking is that if we take the open subset $(-1;1)$ of $\mathbb{R}$ for each $f$ since the map is continuous we get open sets containing $p_i$, and it boils down to showing the intersection is empty. I can't quite follow this important part through.
 A: If $q\ne p$, let $f_p$ be a function $f:~X\longrightarrow\mathbb{R}$ satisfying $f^{-1}(0)=\{p\}$ (this exists by assumption), and let $\alpha=f_p(q)>0$, and let $\epsilon=\frac{\alpha}2$. Consider the inverse images under $f_p$ of $(\leftarrow,\epsilon)$ and $(\epsilon,\to)$.
A: For each continuous function $f\colon X\to \mathbb R$, define $f^{\times}$ on $X\times X$ by $f^{\times}(p,q)=f(p)-f(q)$. 
Consider the intersection: $$\bigcap_{f\colon X\to\mathbb R}\left(f^{\times}\right)^{-1}(\{0\})$$
A: Let $p,q$ be distinct points in $X$.
Suppose there exists continuous $f: X \rightarrow \mathbb{R}$ such that $f^{-1}(0)= \{ p \}$. Notice, for some $a>0$, $f^{-1}(-a,a) = U \subseteq X$ and $U$ is open in $X$ as it is the inverse image of an open set of real numbers. Of course, $p \in U$ is also clear from its construction. 
Likewise, we have the existence of continuous $g: X \rightarrow \mathbb{R}$ for which $g^{-1}(0) = \{ q \}$. Moreover, for some $b>0$, $g^{-1}(-b,b) = V \subseteq X$ has $V$ open containing $q$.
So, there you have it, two open sets, each containing one of the points of interest. Now, we just need (possibly) to find smaller sets which are disjoint. How to do that?
