$\mathcal L$ is not connected or Hausdorff, but if you delete its countably
many isolated points the result is a closed, connected, convex sublattice (which is
still not Hausdorff).

To see this, call a function $f\in \mathbb N^{\mathbb N}$ *passive* at $n$ if
$f(n+1)=f(n)$, and *aggressive* at $n$ if $f(n+1)=f(n)+1$. If $\mathcal L_0$
is the sublattice of $\langle \mathbb N^{\mathbb N}, \max, \min\rangle$
consisting of functions either passive or aggressive at each $n$, then
$\mathcal L$ is the quotient of $\mathcal L_0$ modulo the equivalence relation
of eventual equality.

If a function is passive at almost all $n$, then it is
equivalent to a unique constant function. The set of these is
an $\omega$-chain at the bottom of $\mathcal L$: all other elements of $\mathcal L$
lie above this $\omega$-chain.

If a function is aggressive at almost all $n$, then it is equivalent to
a unique function of the form $f_k(n) = n+k$ where $k\in \mathbb Z$.
Here, when $k\geq 0$, I really mean $f_k(n) = n+k\in \mathbb N^{\mathbb N}$,
but for negative subscripts what I mean is
$f_{-k}(n) = \max\{0, n-k\}\in \mathbb N^{\mathbb N}$.
In any case, the equivalence classes of functions
$f_k(n)$ are ordered the same way the subscripts are, and this
yields a $\mathbb Z$-chain at the top of $\mathcal L$. All
elements of $\mathcal L$ not in this $\mathbb Z$-chain lie below
the entire chain.

All of this shows that $\mathcal L$ looks like the ordinal sum
$\mathbb N + C + \mathbb Z$, where $C$ is the ``core'' of
passive-aggressive functions, i.e. those that are passive
infinitely often and aggressive infinitely often.

The chains at the top and bottom of $\mathcal L$
consist of points isolated in the
interval topology, so of course $\mathcal L$ is not connected.
I want to argue that $C$, the passive-aggressive core of $\mathcal L$,
is connected but not Hausdorff.
($C$ is the intersection of all closed intervals of the form
$[c,f_k(n)]$, where $c$ denotes an equivalence class represented by a constant function, so $C$ is closed, convex, and the subspace
topology on $C$ is the interval topology on $C$.)

The assertions about $C$ follow from:

**Claim.** $C$ is not equal to a finite union of proper
closed intervals.

*Proof of Claim.*
Here a closed interval in $C$ has the form $[f):=[f,\infty)$,
$(g]:=(-\infty,g]$ or $[f,g]$. (I will use $f$ and $g$ to denote
functions and also their equivalence classes in $\mathcal L$.)
For the claim we can restrict
attention to intervals of the first two types, since if
$[f,g]$ is proper, then one of the larger intervals $[f)$
or $(g]$ is proper, and we can use that in place of $[f,g]$
in our union.

So assume that $C = [f_1)\cup \cdots \cup [f_r)\cup (g_1]\cup\cdots\cup (g_s]$
where each interval is proper. Since
the functions $f_i, g_j$ represent elements of $C$
they are passive-aggressive.
We build a passive-aggressive function $h$ that is
not in the union. Start by defining $h$ at $0, 1, 2, \ldots$
so that it is aggressive at every $n$ until we reach a value
$n_0$ where $h(n_0)$ strictly majorizes every $g_j(n_0)$. This is possible,
since the $g_j$'s are passive infinitely often. Now continue
defining $h$ so that it is passive at $n_0+1, \ldots, n_1$ until $h(n_1)$
is strictly majorized by every $f_i(n_1)$. This is possible,
since the $f_i$'s are aggressive infinitely often. Continue
this, alternating back and forth, to construct an $h$ that strictly majorizes
every $g_j$ infinitely often and which is strictly majorized by every $f_i$
infinitely often. This $h$ is passive-aggressive, but
does not represent an element of
$[f_1)\cup \cdots \cup [f_r)\cup (g_1]\cup \cdots\cup (g_s]$. \\

**Corollary.**
$C$ has no pair of disjoint nonempty open sets.

*Proof of Cor:*
If $U$ and $V$ are such, then we may assume they
are nonempty, basic, open sets. The complements
$C\setminus U$ and $C\setminus V$ are proper, basic, closed sets
whose union is $C$. But a basic closed set is a finite union
of closed intervals, so $(C\setminus U) \cup (C\setminus V)$
now expresses $C$ as a finite union of proper closed intervals,
contrary to the Claim. \\

But if $C$ has no pair of disjoint nonempty open sets,
it is connected and not Hausdorff.