My question comes from a demonstration made by my professor.
Let $\Delta \subseteq M \times M$ be the diagonal where $\Delta := \{(x,x) | x \in M\}$, where $M$ is a $\text{metric space}$. Consider $d$ to be the $\text{metric}$ $d(\mathbf{x},\mathbf{y}) = \max\{d(x_1,y_1),d(x_2,y_2)\}$. Show that $d(\mathbf{x},\Delta) = 0 \Leftrightarrow \mathbf{x} \in \Delta$.
His proof went as following for the return, where I didn't understand.
$(\Leftarrow):$ Suppose $\mathbf{x} \notin \Delta$. Then $x_1 \neq x_2$ and $d(x_1,x_2) = c > 0$ Take $\mathbf{p}=(p,p) \in \Delta$ anywhere. We have:
$d(\mathbf{x},\mathbf{p}) = \max\{d(x_1,p),d(x_2,p)\} \geq d(x_i,p),\quad i = 1,2\quad \Longrightarrow$
$d(\mathbf{x},\mathbf{p}) \geq \frac{d(x_1,p) + d(x_2,p)}{2} \geq \frac{d(x_1,x_2)}{2} = \frac{c}{2}\quad$
As $d(\mathbf{x},\mathbf{p}) \geq c/2$ for all $\mathbf{p} \in \Delta$, passing for the infimum, we obtain $d(\mathbf{x},\Delta) \geq c/2 > 0$.
My problem here is when he assumes the infimum for the $c$ to be necessarily different from $0$. This seems quite intuitevily clear for this case but for many cases such as $p \in \mathbb{(R\setminus\mathbb{Q})}, \quad d(p,\mathbb{Q}) = 0$. The distance from all points is greater than $0$, but nonethless the $d(p, \mathbb{Q})$ is zero since there is always a rational as close as you want (I know about the property that the rational are dense in the reals). My question still comes from his proof and how he can assume this with confidence.
I tried to translate his proof from Portuguese, so sorry if I made any technical mistake previously...