Proof. (H)$\implies$(H1): If $x\ne y$ then $\exists U,V$ open neighbourhoods of $x,y$ resp such that $U\cap V=\emp$; hence $y\nin\cl{U}$
(H1)$\implies$(H2): Let $y\ne x$ then there exists a closed nbd $V$ of $x$ such that $y\nin V$, and by hypothesis there exists $M\in\frak{F}$ such that $M\suf V$; thus $M\cap V^{c}=\emp$. But $V^{c}$ is a nbd of $y$; hence $y$ is not a cluster point of $\frak{F}$
(H2)$\implies$(H3): Clear, since every limit point of a filter is also a cluster point.
(H3)$\implies$(H4): Suppose the diagonal $\la$ is not closed then $\exists (x,y)\in \cl{\la}\del \la$ such that $x\ne y$. Since $(x,y)\in\cl{\la}$ every basis element $U\ti V$ of $(x,y)$ meets $\la$ but $(U\ti V)\cap \la\ne\emp\iff U\cap V\ne \emp$. Now consider $\frak{B}=\curl{U\cap V:U\in\cali{N}(x),V\in\cali{N}(y)}$ we see from above that $\frak{B}$ is a non-empty filter base generating some filter $\frak{F}$. It must be that $U\cap V\in \frak{F}$ for fixed $U$ and any $V$ thus $U\in \frak{F}$ so $\cali{N}(x)\suf \frak{F}$ that is to say $\frak{F}\to x$. Similarly $\frak{F}\to y$, a contradiction.
(H4)$\implies$(H5): Obvious
(5)$\implies$(H): If $x\ne y$ then $(x,y)\in X\ti X$ is not in the diagonal $\la$, hence there is a neighbourhood $V$ of $x$ and a neighbourhood $W$ of $y$ in $X$ such that $(V\ti W)\cap \la=\emp$, which means that $V\cap W=\emp$.$\square$