My example is, $f : \mathbb{R}^+ \to \mathbb{R}$ defined by: $$f(x) = \begin{cases} x, &\text{if }0 \leq x < 1 \\ \tfrac{1}{x}, &\text{if }x \geq 1. \end{cases}$$ Even though $f(0)=0$ but $f([5,6]) \neq 1$. But $\mathbb{R}^+$ is a Hausdorff space but this continuous function fails to separate the point $0$ and $[5,6]$.
Is this correct ?
On
$\pi$-Base is an online database of topological spaces inspired by Steen and Seebach's Counterexamples in Topology. It lists the following Hausdorff spaces that are not completely regular. You can learn more about the spaces by view the search result.
Alexandroff Plank
An Altered Long Line
Arens Square
Countable Complement Extension Topology
Deleted Diameter Topology
Deleted Radius Topology
Deleted Tychonoff Corkscrew
Double Origin Topology
Gustin's Sequence Space
Half-Disc Topology
Indiscrete Irrational Extension of $\mathbb{R}$
Indiscrete Rational Extension of $\mathbb{R}$
Irrational Slope Topology
Irregular Lattice Topology
Minimal Hausdorff Topology
Pointed Irrational Extension of $\mathbb{R}$
Pointed Rational Extension of $\mathbb{R}$
Prime Integer Topology
Rational Extension in the Plane
Relatively Prime Integer Topology
Roy's Lattice Space
Roy's Lattice Subspace
Simplified Arens Square
Smirnov's Deleted Sequence Topology
Strong Parallel Line Topology
Strong Ultrafilter Topology
A simple example of a Hausdorff space which is not (completely) regular is the K-topology on $\mathbb{R}$ using the set $K = \{ \frac 1n : n \geq 1 \}$. This is formed by taking as a base all open sets in the usual topology, as well as all sets of the form $(a,b) \setminus K$.
Since it is a finer topology, it is Hausdorff.
It is not too difficult to show that $K$ is a closed set in the new topology, and there are no disjoint open set $U,V$ with $0 \in U$ and $K \subseteq V$.
(Even finer grained examples of Hausdorff but not (completely) regular spaces may be found in the answers to my question here. It is easy to show that every regular (Hausdorff) space satisfies the criteria I was asking about.)