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A complete lattice has every join, hence every directed and finite join. Conversely, if a partially ordered set is directed-complete and has finite joins including the empty join, then for any subset $A$ the set of joins of its finite subsets is nonempty and directed. Its least upper bound is $\bigvee A$. Including the empty join supplies a bottom element and handles $A=\varnothing$.

For a directed set of partial functions from $A$ to $B$, their union is a function: any two functions have a common extension, so they agree wherever their domains overlap. This union is exactly their least upper bound in the extension order, proving directed completeness.

Write $C_x$ for the intersection of all sets containing $x$ and closed under $f$ and directed joins. Then $xRy$ means $y\in C_x$. Reflexivity is immediate, and transitivity follows because any closed set containing $x$ must contain $y$ and then every $z$ with $yRz$. The upper set $[x,\infty )$ is closed: $f$ is inflationary, and a directed join of elements above $x$ is above $x$. Thus $xRy$ implies $x\le y$, proving antisymmetry.

If $h,k\in H$, then $xRk(x)Rh(k(x))$, so $h\circ k\in H$. Moreover $h(k(x))\ge h(x)$ by monotonicity, and $h(k(x))\ge k(x)$ by inflationarity. This proves that $Hx$ is directed. Define $h_0(x)=\bigvee Hx$. Pointwise monotonicity of every $h$ implies monotonicity of $h_0$. Since $Hx\subseteq C_x$ and $C_x$ is closed under directed joins, $xR{h}_0(x)$; hence $h_0\in H$.

The original map $f$ belongs to $H$ by the definition of closedness, so $f\circ h_0\in H$ and $f(h_0(x))\le h_0(x)$ by the definition of the join. Inflationarity gives the reverse inequality. Finally, if $p=f(p)$ and $p\ge x$, the lower set $(-\infty ,p]$ is closed by monotonicity of $f$ and the join property. Thus every $h(x)$ is at most $p$, and so is $h_0(x)$. This proves the least fixed point from a directed family of maps: $h_0(x)$ is the least fixed point above $x$.