Let $(E, \mathcal{T})$ be a compact Hausdorff space. It is well known that every topology $\mathcal{U}$ coarser than $\mathcal{T}$ such that $(E, \mathcal{U})$ is Hausdorff is equal to $\mathcal{T}$.
Is the converse true?
(that is: if $\mathcal{T}$ is a coarsest topology amongst Hausdorff topology on $E$, then $(E, \mathcal{T})$ is compact)
Thanks in advance.
If $A$ is the linearly ordered set $\{1,2,3,...,\omega,...,-3,-2,-1\}$ with the order topology, and if $\Bbb Z^+$ is the set of positive integers with the discrete topology, we define $X=A\times\Bbb Z^+$ together with two ideal points $a$ and $-a$. The topology $\tau $ on $X$ is determined by the product topology on $A\times \Bbb Z^+$ together with the local base
$$M_n^+(a)=\{a\} \cup\{(i, j) \mid i<\omega, j>n\} \text{ and } M_{n}^-(-a)=\{-a\} \cup\{(i, j) \mid i>\omega,j<n\} $$
Visualizing $X$:
A straightforward consideration of cases shows that $X$ is Hausdorff.
The collection of all basis neighborhoods form an open covering of $X$ with no finite subcovering (Hint: Consider the points $(\omega,j)$ for each $j\in \Bbb Z^+$). Thus $X$ is not compact.
$X$ is almost compact (i.e. each open cover of the space has a finite subcollection the closures of whose member cover the space). This follows from the fact that the closures of any neighborhoods of $a$ and $-a$ contain all but finitely many of the points $(\omega,j)$. A straightforward consideration of cases shows similarly that the complement of any basis neighborhood is also almost compact.
Now suppose $\tau'\subset\tau$, suppose $N$ is a basis neighborhood of $\tau$ and suppose $\{O_\alpha\}$ is a $\tau'$-open covering of $X \setminus N$. It is then a $\tau$ open covering, so there exist finitely many sets $O_1, O_2,..., O_n$ the union of whose $\tau$-closures covers $X \setminus N$. But the $\text{cl}_{\tau} (O_i)\subset\text{cl}_{\tau'} (O_i)$, so $X \setminus N$ is covered by the union of the $\tau'$-closures of $0_1, O_2, ..., O_n$. In otherwords, $X\setminus N$ is an almost compact subset of $(X,\tau')$
Suppose $\tau'$ is a proper subtopology of $\tau$. Then there would be some basis neighborhood $N\in \tau$ for which $X\setminus N$ is not closed in $\tau'$, and so there would be a point $x\in N$ such that $x\in \text{cl}_{\tau'} (X\setminus N)$. Let $\{C_\alpha\}$ be the collection of all $\tau'$-neighborhoods of $x$, and suppose $\{X\setminus \overline{C_\alpha}\}$ covers $X\setminus N$, then there exists $\overline{C_1},\overline{C_2},...,\overline{C_n}$ such that $X\setminus N\subset \bigcup \left(\overline{X\setminus\overline{C_i}}\right)$ (by 4). But $\bigcup\ \left(\overline{X \setminus \overline{C}_{i}}\right)$ is closed, hence it must contain $x$ (as $x\in \text{cl}_{\tau'} (X\setminus N)$). But $\bigcap C_i$ is a neighborhood of $x$, implying $\left(\bigcap C_{i}\right) \cap\left(\bigcup\ \left(\overline{X \setminus \overline{C}_{i}}\right)\right)=\left(\bigcap C_{i}\right) \cap\left(\overline{\bigcup\ \left(X \setminus \overline{C}_{i}\right)}\right)$ is nonempty, which is clearly impossible. Hence $\{X\setminus \overline{C_\alpha}\}$ doesn't cover $X\setminus N$, which implies there exists $y\notin N$ such that $y\in \bigcap \overline{C_\alpha}$. Since $y\neq x$ are inseparable by open sets, $(X,\tau')$ is not Hausdorff.
In conclusion, $X$ is minimal Hausdorff but not compact.
The above is essentially my paraphrasing of Steen and Seebach's Counterexample in Topology example 100.
