 Solution

Write $h(A,B)={\sup }_{a\in A}\mathrm{dist}(a,B)$, so that $\rho (A,B)=\max \{h(A,B),h(B,A)\}$ is the Hausdorff distance. Since $X$ is a bounded set, there is a finite $M$ with $d(x,y)\le M$. Nonemptiness of $A,B$ therefore gives $0\le h(A,B)\le M$, so $\rho$ is finite. Its symmetry and nonnegativity are immediate, as is $\rho (A,A)=0$.

If $\rho (A,B)=0$, then every $a\in A$ has zero distance to a set $B$. Since $B$ is a closed set, part (a)(iii) gives $A\subseteq B$. Interchanging the sets gives $B\subseteq A$. Thus $\rho (A,B)=0$ implies $A=B$. Closedness is essential here: without it a set and its closure have Hausdorff distance zero.

For the triangle inequality, fix $a\in A$ and $ϵ>0$. The approximation property of an infimum supplies $b\in B$ with $d(a,b)<\mathrm{dist}(a,B)+ϵ$, and then $c\in C$ with $d(b,c)<\mathrm{dist}(b,C)+ϵ$. Consequently,Let $ϵ\downarrow 0$ and take the supremum over $a$ to obtain $h(A,C)\le h(A,B)+h(B,C)$. Applying this inequality in the reverse direction as well yieldsAll metric axioms follow. Boundedness was used to ensure finiteness, and closedness to distinguish different sets; neither was needed for the triangle inequality.