Proving that the lexicographic order is a direction 
Suppose $\Lambda$ is a directed set and $M_{\lambda}$ is a directed set for each $\lambda\in \Lambda$. Show that $\{(\lambda, \mu): \lambda\in\Lambda, \mu\in M_{\lambda}\}$ is a directed set with the relation $\leq$ defined by $$(\lambda_{1}, \mu_{1})\leq (\lambda_{2}, \mu_{2})\iff \lambda_{1} < \lambda_{2}\text{ or }(\lambda_{1} = \lambda_{2}\text{ and }\mu_{1}\leq \mu_{2}).$$

I have included the definition of a directed set for reference.

A relation $\leq$ on a set $\Lambda$ is a direction on $\Lambda$ if it satisfies:
(i) $\lambda\leq\lambda$, for each $\lambda\in\Lambda$.
(ii) If $\lambda_{1}\leq \lambda_{2}$ and $\lambda_{2}\leq \lambda_{3}$ then $\lambda_{1}\leq \lambda_{3}$.
(iii) If $\lambda_{1}, \lambda_{2}\in \Lambda$ then there is some $\lambda_{3}\in\Lambda$ with $\lambda_{1}\leq\lambda_{3}$, $\lambda_{2}\leq \lambda_{3}$.

I am specifically having trouble proving transitivity (property (ii)). Suppose $(\lambda_{1}, \mu_{1})\leq (\lambda_{2}, \mu_{2})$ and $(\lambda_{2}, \mu_{2})\leq (\lambda_{3}, \mu_{3})$ such that $\lambda_{1} < \lambda_{2}$ and $\lambda_{2} < \lambda_{3}$. Since $\Lambda$ is a direction, we have $\lambda_{1}\leq \lambda_{3}$. Is it necessary for the strict inequality $\lambda_{1} < \lambda_{3}$ to hold? If $\lambda_{1} = \lambda_{3}$, I am unable to show that $\mu_{1}\leq \mu_{3}$. Note that the relations on $\Lambda$ and each $M_{\lambda}$ need not be antisymmetric.
I believe I am missing something trivial here. Any help would be greatly appreciated. Thanks in advance.
Context: I am solving the following problem in Willard's general topology textbook:

If $x_{\lambda}\rightarrow x$ and for each $\lambda\in\Lambda$, a net $(x_{\mu}^{\lambda})_{\mu\in M_{\lambda}}$ converges to $x_{\lambda}$ then there is a diagonal net converging to $x$.

 A: I think I have a counterexample. Let
$$\Lambda=\{S,T\}\text{ with }S\le T\text{ and }T\le S$$
$$M_S=\{a,b,c\} \text{ with } a\le c\text{ and }b\le c$$
$$M_T=\{d\}$$
(I have omitted the reflexive elements of these relations for clarity). Then we have $(S,a)\le(T,d)\le(S,b)$, but not $(S,a)\le(S,b)$.
A: Willard, I think, means the order as defined by Kelley in his book (a pioneer in the use of directed sets and nets in topology), in that the set is $\Lambda \times \prod_{\lambda \in \Lambda} M_\lambda$ where the direction on $\prod_{\lambda \in \Lambda}M _\lambda$ is defined coordinatewise: $(m_\lambda) \le (n_\lambda)$ iff $\forall \lambda \in \Lambda: m_\lambda \le n_\lambda$. And finally $(\lambda_1, (m_\lambda)) \le (\lambda_2, (n_\lambda))$ iff $\lambda_1 \le \lambda_2$ and $(m_\lambda) \le (n_\lambda)$ etc. This is not "lexicographic" but at least totally unproblematic and has the required property for the diagonal net (when we map $(\lambda, (m_\lambda))$ to $x^\lambda_{m_\lambda}$ etc.).
Engelking 1.6.B does it the same way.
