Prove that $(S, d)$ is a metric space, i.e. $d$ is a distance function on $S$ . Let $S$ be the set of nonempty compact subsets of $\mathbb{R}^{2}$.
For any $r > 0$ and $K ∈ S$ , we define the $r -neighborhood$ of $K$ to be
$$B_{r}(K) := \{x ∈ \mathbb{R}^2: d(x, a) < r\; for \; some \;a ∈ K\}=\bigcup_{a\in K}B_{r}(a)$$
For $K_{1}$, $K_{2}\in S$, we define
$$d(K_{1},K_{2}):=\inf \{ r>0:K_{1} \subset B_{r}(K_{2}) \; and \; K_{2} \subset B_{r}(K_{1}) \}$$
Prove that $(S,d)$ is a metric space.
As we always do, I proved with the definition of a Metric Space.

*

*Def.1: Consider $d(K_{a},K_{a})$ & $d(K_{a},K_{b})$ for all $K_{a} \neq K_{b}$
$$d(K_{a},K_{a})$$ $$= \inf \{r>0: K_{a} \subset B_{r}(K_{a}) \}$$ $$= \inf \{ r:r>0 \}=0$$ $$\Rightarrow d(K_{a},K_{a})=0 \; (\forall K_{a} \in S)$$
And, $$d(K_{a},K_{b})$$ $$= \inf \{r>0: K_{a} \subset B_{r}(K_{b}) \;and\;  K_{b} \subset B_{r}(K_{a}) \}$$
Since $K_{a} \neq K_{b}$, $\exists x \in K_{a}$, but $x \notin K_{b}$.
Then, $\exists y \in K_{b}$ s.t $d_{\mathbb{R}^2}(x,y)$ is least possible.
Then, $\inf \{r>0: K_{a} \subset B_{r}(K_{b}) \;and\;  K_{b} \subset B_{r}(K_{a}) \} \geq d_{\mathbb{R}^2}(x,y)>0$
Thus, $d(K_{a},K_{b})>0$
Hence, Def.1. satisfies for $(S,d)$


*Def.2.: Consider $d(K_{a},K_{b})$ $\; \forall K_{a},K_{b} \in S$
$$d(K_{a},K_{b}) = \inf \{r>0: K_{a} \subset B_{r}(K_{b}) \;and\;  K_{b} \subset B_{r}(K_{a}) \}$$
If $K_{a}=K_{b}$, then $d(K_{a},K_{b})=0=d(K_{b},K_{a})$
If $K_{a} \neq K_{b}$, then $$d(K_{a},K_{b})$$ $$= \inf \{r>0: K_{a} \subset B_{r}(K_{b}) \;and\;  K_{b} \subset B_{r}(K_{a}) \}$$ $$ = \inf \{r>0: K_{b} \subset B_{r}(K_{a}) \;and\;  K_{a} \subset B_{r}(K_{b}) \}$$ $$=d(K_{b},K_{a})$$
Hence, Def.2. satisfies for $(S,d)$


*Def.3.: Consider $d(K_{a},K_{c})$ $\; \forall K_{a},K_{c} \in S$
If $K_{a}=K_{c}$, then $d(K_{a},K_{c})=0\leq d(K_{a},K_{b}) + d(K_{b},K_{c})$ $\; \forall K_{b} \in S$ $\; \Rightarrow d(K_{a},K_{c}) \leq d(K_{a},K_{b}) + d(K_{b},K_{c})$
If $K_{a} \neq K_{c}$, then $\exists x \in K_{a}$, but $x \notin K_{c}$.
Also, $\exists y \in K_{c}$ s.t $d_{\mathbb{R}^2}(x,y)$ is greatest possible.
Let $z \in K_{b}$ s.t $d_{\mathbb{R}^2}(x,z),d_{\mathbb{R}^2}(z,y)$ are less than or equal to $d(K_{a},K_{b}),d(K_{b},K_{c})$
Note that, $\inf \{r>0: K_{a} \subset B_{r}(K_{c}) \;and\;  K_{c} \subset B_{r}(K_{a}) \} \leq d_{\mathbb{R}^2}(x,y)$ $\Leftrightarrow d(K_{a},K_{c})\leq d_{\mathbb{R}^2}(x,y)$
Also, by Triangle Inequality, $d_{\mathbb{R}^2}(x,y) \leq d_{\mathbb{R}^2}(x,z) + d_{\mathbb{R}^2}(z,y)$ $$\Rightarrow d(K_{a},K_{c}) \leq d_{\mathbb{R}^2}(x,y) \leq d_{\mathbb{R}^2}(x,z) + d_{\mathbb{R}^2}(z,y) \leq d(K_{a},K_{b}) + d(K_{b},K_{c})$$
Hence, Def.3. satisfies for $(S,d)$
Consequently, $(S,d)$ is a metric space. $$\blacksquare$$
Is it Okay to prove in this way? Is it clear enough?
 A: There are some specific points that need attention:

Since $K_{a} \neq K_{b}$, $\exists x \in K_{a}$, but $x \notin K_{b}$.

This is not true when $K_a$ is a subset of $K_b$, and if $K_a$ is a strict subset of $K_b$ then $K_a \neq K_b$ holds as well. But, since $K_a \neq K_b$, then one of the sets must not be a subset of the other; either $K_a \not\subseteq K_b$ or $K_b \not\subseteq K_a$ holds. So, you can assume $x \in K_a \setminus K_b$ exists, or $x \in K_b \setminus K_a$ exists.
Of course, if you prove symmetry of $d$ first, then you can assume without loss of generality that $x \in K_a \setminus K_b$ exists, as the roles of $K_a$ and $K_b$ are now symmetric.
You also repeat this error when it comes to proving triangle inequality, but again, the roles of $K_a$ and $K_c$ are symmetric if you already have symmetry of $d$ proven.

Then, $\exists y \in K_{b}$ s.t $d_{\mathbb{R}^2}(x,y)$ is least possible.

This is very much true, but it might be worth citing a theorem to back this up, or at least mentioning that $d_{\Bbb{R}^2}(x, \cdot\,)$ is a continuous function, and thus achieves a minimum on the compact set $K_b$.

Def.2.: Consider $d(K_{a},K_{b})$ $\; \forall K_{a},K_{b} \in S$
$$d(K_{a},K_{b}) = \inf \{r>0: K_{a} \subset B_{r}(K_{b}) \;and\;  K_{b} \subset B_{r}(K_{a}) \}$$
If $K_{a}=K_{b}$, then $d(K_{a},K_{b})=0=d(K_{b},K_{a})$

This is fine, but you shouldn't need to treat the $K_a = K_b$ case separately. The definition of $d$ makes it pretty clear that the roles of $K_1$ and $K_2$ are interchangeable, and indeed the argument following does not use the assumption that $K_a \neq K_b$.

Let $z \in K_{b}$ s.t $d_{\mathbb{R}^2}(x,z),d_{\mathbb{R}^2}(z,y)$ are less than or equal to $d(K_{a},K_{b}),d(K_{b},K_{c})$

Such a $z$ may not exist. Let $K_a, K_b, K_c$ be closed unit balls centred at $(-1, 0), (0, 0), (0, 1)$ respectively. If we take $x \in K_a$ to be $(-2, 0)$, then by your construction, we must take $y$ to be the furthest point in $K_c$ from $x$, which is $(2, 0)$. We can then compute $d(K_a, K_b) = d(K_b, K_c) = 1$, and so we want $z$ such that
\begin{align*}
d_{\Bbb{R}^2}(z, x) &\le d(K_a, K_b) = 1 \\
d_{\Bbb{R}^2}(z, y) &\le d(K_b, K_c) = 1.
\end{align*}
Given $d_{\Bbb{R}^2}(x, y) = 4$, this would violate triangle inequality on $\Bbb{R}^2$.
This would seem to be a fatal error to your approach with the triangle inequality. You will need to come up with another approach.
P.S. This metric has a name: the Hausdorff metric.
