Definition: A function $$\delta: X\times 2^X\to [0,\infty]$$ with $X$ called approach space if it satisfies the following properties for all $x\in X$ and $A,B\subset X$
i) $\delta(x,\{x\})=0$
ii) $\delta(x,\emptyset)=\infty $
iii) $\delta(x,A\cup B)=\min\{\delta(x,A),\delta(x,B) \}$
iv) $\delta(x,A)\leq \delta(x,A^\varepsilon)+\varepsilon, \quad A^\varepsilon=\{a\in A: \delta(a,A)\leq \varepsilon \}$.
I'm trying prove for every $x\in X$ and $A,B\subset X$ $$\delta(x,A)\leq \delta(x,B)+\sup_{b\in B} \delta(b,A).$$
My proof:
Suppose for some $x\in X$ the inequality isn't hold, namely
$$\delta(x,A)> \delta(x,B)+\sup_{b\in B} \delta(b,A).$$
If we take $x:=b$, then \begin{align*} \delta(b,A)& > \delta(b,B)+\sup_{b\in B} \delta(b,A)\\ & > \delta(b,B)+\delta(b,A).\\ \end{align*} This leads to a contradiction ($\delta(b,B)<0$).
Are my proof steps correct? Thanks in advance.