When working with the Hausdorff distance, its better to work with a metric space where $d\le 1$. In principle, this is not a problem, since every metirc space is homeomorphic to one where the metric is bounded.
Given this modification, we can define, for arbitrary subsets of $X$,
$$
h(A,B):=\begin{cases}
0 & \text{if } A=B=\emptyset\\
1 & \text{if exactly one of $A$ and $B$ is $\emptyset$}\\
\max\left\{\sup_{a\in A}r(a,B), \sup_{b\in B}r(A,b)\right\}& \text{if }A\neq\emptyset \wedge B\neq\emptyset
\end{cases}
$$
Then you can show that $h$ is a pseudo metric on $\mathcal{P}(X)$, the key being the inequality:
$$
d(x,Y)≤d(x,Z)+h(Y,Z)
$$
where we define $d(x,\emptyset)=1$.
You can also prove that $h(A,B)=0\iff \bar A=\bar B$, which shows that $h$ is a metric on the family of closed sets.
When dealing with compact sets, you get easier proofs of the facts above, together with many extra properties. See I.4.F in Kechris' Classical Descriptive Set Theory
Yes, you can:
Define (for convenience's sake) $$ \rho(A,B) = \sup_{a \in A} d(a,B) \text{ and } \rho(B,A) = \sup_{b \in B} d(b,A)$$
so that $$=d_H(A,B)=\max(\rho(A,B),\rho(B,A))$$
I claim that for $\rho$ we also have a triangle inequality:
$$\rho(A,B) \le \rho(A,C) + \rho(C,B)\tag{0}$$ for compact, non-empty $A,B,C \subseteq X$.
To see that: let $a \in A$ be arbitrary. By compactness of $C$, there is some $c_a \in C$ so that $d(a,c_a)=d(a,C)$.
Then: $$d(a,B)= \inf_{b \in B} d(a,b) \le \inf_{b \in B} \left(d(a,c_a) + d(c_a,b)\right)$$ by the standard triangle inequality for $d$ ($a$ to $b$ via $c_a$). The left summand does not depend on $b$ so continuing the previous:
$$\inf_{b \in B} \left(d(a,c_a) + d(c_a,b)\right) = d(a,c_a) + \inf_{ \in B} d(c_a,B) = d(a,C) + d(c_a,B) \le \rho(A,C) + \rho(C,B)$$ by the definition, choice of $c_a$ and using that the $\rho$ is a sup, so upperbound. The right hand side is a fixed number which is an upperbound for $d(a,B)$, while $a\in A$ was arbitrary, so $(0)$ follows, as the sup is is the smallest upperbound.
Now the triangle inequality for $d_H$ is an easy consequence, using $x \le \max(x,y)$ etc: first note
$$\rho(A,B) \le \rho(A,C) + \rho(C,B) \le d_H(A,C) + d_H(C,B)\tag{a}$$
and then
$$\rho(B,A) \le \rho(B,C) + \rho(C,A) \le d_H(C,B) + d_H(C,A)\tag{b}$$
The right hand sides of $(a)$ and $(b)$ are identical by symmetry of $d_H$, and as both $\rho(A,B)$ and $\rho(B,A)$ are upper-bounded by it, so is their maximum, so
$$d_H(A,B) \le d_H(A,C) + d_H(C,B)$$ as required. QED.
Best Answer
Let $\epsilon,\epsilon'>0$ be such that $$A\subseteq U_\epsilon (B), \quad B\subseteq U_\epsilon(A);$$ $$B\subseteq U_{\epsilon'} (C), \quad C\subseteq U_{\epsilon'}(B).$$ Note that if $x\in U_\epsilon (B)$, then by definition there exists $b\in B$ such that $d(x,b)<\epsilon$. As $b\in B\subseteq U_{\epsilon'} (C)$, there exists $c\in C$ such that $d(b,c)<\epsilon'$. It follows that $$d(x,c)\leq d(x,b)+d(b,c)<\epsilon+\epsilon'\quad\Rightarrow\quad A\subseteq U_{\epsilon+\epsilon'}(C).$$ Similarly we can show that $C\subseteq U_{\epsilon+\epsilon'}(A)$. Therefore we have $$\begin{aligned}&\{ \epsilon >0 \mid A\subset U_\epsilon (B) \text{ and } B\subset U_\epsilon(A)\}+\{ \epsilon' >0 \mid B\subset U_{\epsilon'} (C) \text{ and } C\subset U_{\epsilon'}(B)\}\\ \subseteq&\{ \delta >0 \mid A\subset U_\delta (C) \text{ and } C\subset U_\delta(A)\}.\end{aligned}$$ Take infimum on both sides yields $$d_H(A,B)+d_H(B,C)\geq d_H(A,C).$$