I can't understand the conclusion of the proof of theorem 43.7 in Munkres' Topology.
We consider a metric space $(X, d)$ and a fixed pont $x_0 \in X$. He defines an imbedding $h : X \to \mathcal{B}(X, \mathbb{R})$ setting $\Phi(a)=\phi_a: X \to \mathbb{R}$ such that $\phi_a(x)=d(x, a)-d(x, x_0)$. He proves that $h$ is an isometrical imbedding, meaning \begin{equation} \rho (\phi_a, \phi_b)=d(a, b) \end{equation} for all $a, b$ in $X$. I understand the argument which leads to the conclusion that \begin{equation} \rho (\phi_a, \phi_b)\le d(a, b) \end{equation} but I can't figure out how the reverse inequality is proved. Munkres writes:
On the other hand, this inequality cannot be strict, for when $x=a$ \begin{equation} |d(x, a)- d(x, b)|=d(a, b). \end{equation}
What does this prove? To me, this just implies that we cannot replace $\le$ with $<$; not that equality holds.
Edit: Is there a particular reason why my question has been downvoted? I think I have provided adequate context (and a precise reference). It lacks an attempt of self-answer just because it is about a specific passage of a proof which I can't understand (in fact I've tagged the question as proof explanation).
I suppose $\rho$ stands for the supremum metric. We have $\phi_a(x)-\phi_b(x)=d(x,a)-d(x,b)$. So $\rho (\phi_a-\phi_b) \geq |(\phi_a-\phi_b) (a)|=d(a,b)$. I just used the fact that the supremum of a set is greater than or equal to any element of the set.