Question

Discrete order topology does not have least upper bound property

Answer 1

It's an exercise in definitions:

Suppose that $\mathcal{T}$ is any topology that contains both the upper topology $\mathcal{T}_u$ and lower topology $\mathcal{T}_l$ on $X$.

Let $x \in X$, then $(\leftarrow,x] \in \mathcal{T}_l \subseteq \mathcal{T}$ and also $[x,\rightarrow) \in \mathcal{T}_u \subseteq \mathcal{T}$.

So, as topologies are closed under finite intersections,

$$\{x\} = (\leftarrow, x] \cap [x, \rightarrow) \in \mathcal{T}$$

and as $x \in X$ was arbitrary $\mathcal{T}$ is the discrete topology (every subset is a union of singletons etc.).

So $\mathcal{T}_l$ and $\mathcal{T}_u$ have a unique common upper bound, the discrete topology, which dierctly implies your first statement.

Now suppose $(X,\le)$ is directed (on the right). Suppose that $\mathcal{T}$ is a common lower bound (in inclusion) for $\mathcal{T}_l$ and $\mathcal{T}_u$. Suppose $O \neq \emptyset$ is an open subset in $\mathcal{T}$. Then let $p \in O$. Then $O$ being open in $\mathcal{T}_u$ implies that $[p, \rightarrow) \subseteq O$. If now $x \in X$ is arbitary, the right-directedness of $X$ implies we have some $x'\in X$ with $p \le x'$ and $x \le x'$. Hence $x' \in O$ (as $x \in [p, \rightarrow)$) and, using that $x' \in O$ and $O$ open and so left-open too, $x \in O$. As $x \in X$ was arbitrary, any non-empty subset in $\mathcal{T}$ is $X$, or $\mathcal{T} = \{\emptyset,X\}$, the trivial topology (i.e. the coarsest topology on $X$). Now adapt this argument slightly to cover the left directed case too.