By triangle inequality you have:
$$|\rho (x_1,y_1)-\rho (x_2,y_2)| \leq |\rho (x_1,y_1)-\rho (x_2,y_1)| + |\rho (x_2,y_1)-\rho (x_2,y_2)| \; ,$$
hence, by reverse triangle inequality:
$$|\rho (x_1,y_1)-\rho (x_2,y_2)| \leq \rho (x_1,x_2) + \rho(y_1,y_2)\; ;$$
from the definition of $\rho\times \rho$ follows:
$$|\rho (x_1,y_1)-\rho (x_2,y_2)| \leq 2(\rho\times \rho) \big( (x_1,x_2) ,(y_1,y_2)\big) \; ,$$
therefore $\rho (x,y)$ is actually Lipschitz.
Intuitively, the plane $\Bbb R^2$ in the lexicographic order, is just a set of vertical lines, all of which are topological copies of $\Bbb R$ and such that different vertical lines do not really interact: the horizonal line $\mathbb{R} \times \{0\}$ has the discrete topology as $(x,0)$ has the neighbourhood $O(x):= \{x\} \times (-1,1)$ which equals the open interval $((x,-1),(x,1))$ in the order, such that $O(x) \cap \left(\mathbb{R} \times \{0\}\right) = \{(x,0)\}$. Some more thought will tell you that in fact $\Bbb R^2$ in the order topology is just homeomorphic to the product $(\Bbb R, \mathcal{T}_d) \times (\Bbb R, \mathcal{T}_e)$, where the first factor has the (metrisable) discrete topology and the second the normal Euclidean one. This suggests a metric: we can use the sum (or $\max$-metric) of the component metrics of the factors to get the right product topology, so define
$$d((x_1, y_1), (x_2,y_2)) = \begin{cases}
|x_2-y_2| & \text{ if } x_1 = y_1 \\
1+|x_2-y_2| & \text{ if } x_1 \neq y_1 \\
\end{cases}$$
which is then (by standard facts, I hope for you) a metric that induces the product topology of the discrete metric and the Euclidean metric, and so the right topology for the lexicographically ordered plane.
Best Answer
Let's assume $p$ is indeed a metric. Then we have a metric space $(X\times X,p)$ and a function $d:X\times X\to\mathbb R$, which we want to show is continuous with respect to $p$.
Let $(x,y)\in X\times X$ and $\varepsilon>0$. If $(x',y')\in X\times X$ then by definition, $$p((x,y),(x',y'))=d(x,x')+d(y,y').$$ We want to make $|d(x,y)-d(x',y')|<\varepsilon$ by forcing the above to be small enough. If $p((x,y),(x',y'))<\delta$ then we know that both $d(x,x')<\delta$ and $d(y,y')<\delta$. This means that $$d(x,y)\le d(x,x')+d(x',y)<\delta+d(x',y)$$ and $$d(x',y)\le d(x',x)+d(x,y)<\delta+d(x,y)$$ and so $$|d(x,y)-d(x',y)|<\delta.$$ Similarly $|d(x',y')-d(x',y)|<\delta$. Thus, $$|d(x,y)-d(x',y')|\le|d(x,y)-d(x',y)|+|d(x',y)-d(x',y')|<2\delta$$ and we can choose $\delta=\varepsilon/2$.