Metric space properties 
I am doing this question from mainly an analysis viewpoint since I haven't covered metric spaces yet.
I have completed part (i). Now I am stuck on (ii). How do I show it is well defined and satisfies (a),(b) and (c).
 A: For well-defined: you need to check that the function actually outputs a real number.  Since we are taking a $\sup$ its possible that the function could output $\infty$.  But if $K, L$ are a closed bounded set, then what can you conclude about $\sup_{l \in L} d(l,K)$
For (a), First assume $\rho(L,K) = 0$.  Then, both $\sup_{l \in L} d(l,K) = 0$ and $sup_{k \in K} d(k,L)$ must be $0$.  So if we fix an $l \in L$ what must be $d(l,K)$?  Well if the $\sup$ is $0$, we must have $d(l,K) = 0$.  Similarly, for any $k \in K$, $d(k, L) = 0$.  Now , you may need to recall the definition of $d(x, V)$ at this point, but after doing so, you will notice that since $L, K$ are closed $d(l,K) = 0 \implies l \in K$ and similarly $d(k,L) = 0 \implies k \in L$ ($d$ is defined by an $\inf$ so if $d(l,K) = 0$ then we can construct a sequence $\{k_n\}$ in $K$ converging to $l$, by closure, $l \in K$). So we have shown: $k \in K \implies k \in L$ and $l \in L \implies l \in K$.  Hence, $L=K$.
For the other direction, if $L = K$, then it is not hard to see that $\sup_{k \in K} d(k,L) = 0$ and $\sup_{l \in L} d(l,K) = 0$.  Hence, their $\max$ is $0$ also.
(b) Is not difficult since for any real numbers $x,y$, $\max\{x,y\} = \max\{y,x\}$.
(c) First, notice that for fixed $k \in K$ and $m \in M$,
$$d(k,L) = \inf_{l \in L} \|k-l\| = \inf_{l \in L} \|k - m + m -l\| \le \inf_{l \in L} (\|k - m \| + \|m - l\|)$$
Since this is true for any $m$, we can take an $\inf$ over the $n \in M$ to conclude
$$d(k,L) \le \inf_{n \in M} \| k - n \| + \inf_{l \in L} \| m - l \| = d(k, M) + d(m,L)$$
Now, let's take a $\sup$ over the $m \in M$ on both sides
$$d(k,L) = \sup_{m \in M} d(k,L) \le \sup_{m \in M} d(k, M) + d(m,L) = d(k, M) + \sup_{m \in M} d(m,L)$$
and then finally take a $\sup$ over the $k \in K$ on both sides
$$\sup_{k \in K} d(k,L) \le \sup_{k \in K} d(k,M) + \sup_{m \in M} d(m,L)$$
Similarly, one can show
$$\sup_{l \in L} d(l,K) \le \sup_{l \in L} d(l,M) + \sup_{m \in M} d(m,K)$$
Now, let $a = \sup_{k \in K} d(k,M)$ and $b = \sup_{m \in M} d(m,L)$, $d = \sup_{l \in L} d(l,M)$ and $c = \sup_{m \in M} d(m,K)$.  We then have,
$$\rho(L,K) = \max\{ \sup_{k \in K} d(k,L), \sup_{l \in L} d(l,K)\} $$
$$\le  \max\{ a+b, c+d\} \le \max\{ a, c\} + \max\{ b,d\} = \rho (M, K) + \rho(L,M)$$
Where we used the inequality in the hint.
