EDIT: I mentioned in the comments that Nik's list of definitions and theorems looks to me strikingly similar to Wallman's generalization of Stone duality.
After getting some feedback from Nik and taking some time to think about the problem more, the connection seems less direct than I previously suggested. However, I do still feel that there is a strong connection between Wallman's theory and Nik's, and I have edited my post to explain this as best I can.
This post does not completely answer Nik's question, but I hope it provides a useful partial answer.
Wallman's Construction:
The main difference between Wallman's theory and yours is that you seem to be thinking of lattice elements as points, whereas Wallman thought of them as closed sets.
I can't find a clear and thorough account of Wallman's results online except for his original paper (which is a bit lengthy, and uses some outdated terminology)
H. Wallman, ``Lattices and topological spaces," Annals of Mathematics vol. 39 (1938), pp. 112 - 126, available here.
I'll outline the construction here.
To every "compact, Hausdorff" lattice $L$ we may associate a topological space called $W(L)$. The points of $W(L)$ are not the members of $L$, but rather the ultrafilters on $L$. The idea is to put a topology on $W(L)$ in such a way that $L$ becomes (isomorphic to) the lattice of closed subsets of $W(L)$.
To put a topology on $W(L)$, let each $a \in L$ define a closed set
$$C_a = \{p \in W(L) : a \in p\}.$$
In other words, $C_a$ is the set of all ultrafilters containing $a$. With your definition of a compact lattice, it's not hard to check that these sets constitute the closed subsets of a topology on $W(L)$.
(Sidebar: If you don't require that every subset of $L$ has a greatest lower bound, then this construction will still work. Instead of giving you a topology, the sets $C_a$ will then define a basis for the closed sets of a topology. To see this idea at work, I recommend the first part of section 1 of this paper of Dow and Hart.)
Given a closed set $C$ in $W(L)$, we obtain a "closed" subset of $L$ (in your terminology) by considering $\{a \in L : C \subseteq C_a\}$. Given a "closed" subset $C$ of $L$, we obtain a closed subset of $W(L)$ corresponding to it, namely $\bigcap_{a \in C}C_a$.
As Nik points out in the comments, the converse is not necessarily true: in general, the lattice of closed subsets of a compact Hausdorff space is not necessarily a "compact Hausdorff" lattice.
Why I think Wallman's idea is relevant here:
Using Wallman's construction and known facts from point-set topology, we can derive "near-miss" versions of some of Nik's theorems. For example, we can get a weaker version of Theorem 5, where we drop the continuity requirement but still require our map to be order-preserving:
Theorem: Suppose $L$ is a "compact Hausdorff" lattice and $a \not\leq b$ are in $L$. Then there is an order-preserving map from $L$ into $[0,1]$ taking $a$ to $1$ and $b$ to $0$.
Proof: Let $X = W(L)$ be the compact Hausdorff space whose lattice of closed sets is isomorphic to $L$. In $X$, $a$ and $b$ correspond to the nonempty closed sets $C_a$ and $C_b$, and the relation $a \not\leq b$ translates to $C_a \not\subseteq C_b$. Let $x$ be a point of $X$ that is in $C_a$ but not in $C_b$. Since $X$ is $T_{3\frac{1}{2}}$, there is a continuous function $f: X \rightarrow [0,1]$ such that $f(x) = 1$ and $f(C_b) = 0$. Let $\varphi: L \rightarrow [0,1]$ be defined by
$$\varphi(d) = \sup \{f(z) : z \in C_d\}.$$
Clearly $\varphi(a) = 1$ and $\varphi(b) = 0$, and $\varphi$ is order-preserving. However, it is easy to come up with examples where $\varphi$ is does not preserve arbitrary meets and joins (hence the "near-miss").
Similarly, we can get a near-miss version of Theorem 6:
Theorem: If $L$ is a "compact Hausdorff" lattice, then there is an order-preserving injection from $L$ into a power of $[0,1]$.
Proof: Let $I = \{(a,b) \in L \times L : a \not\leq b\}$. For each $(a,b) = i \in I$, fix an order-preserving function $\varphi_i$ as in the previous theorem. Define $\varphi: L \rightarrow [0,1]^I$ by $\pi_i \circ \varphi(a) = \varphi_i(a)$. This function is clearly order-preserving. For injectivity, consider that for all $a \neq b \in L$, either $a \not\leq b$ or $b \not\leq a$.
We get a near-miss version of Theorem 4 in the same way (first prove a $0$-dimensional analogue of Theorem 5, stating that the function in question can be assumed to be two-valued).
Best Answer
If $Z$ is not compact, and $X=\{p\}\cup Z$ is the space whose nonempty open sets are of the form $\{p\}\cup V$ with $V$ open in $Z$, then $X$ is not compact, but every continuous function from $X$ to a Hausdorff space is constant.