We have got $d_r$ metric
$$d_r(x,y) = \begin{cases} |x_2-y_2|, & \text{if $x_1 = y_1$;} \\ |x_2| + |y_2| + |x_1-y_1|, & \text{if $x_1 \neq y_1 $} \end{cases}$$
Prove that inside $(\mathbb{R}^2, d_r)$ space K set is compact $\Leftrightarrow$ K is closed and lies within some set $ [a,b] \times \{0\} \cup \bigcup_{n=1}^{\infty} {s_n} \times [-t_n, t_n]$, where $s_n \in [a,b]$ and $t_n \rightarrow 0$
I have got some idea.
First show, that if K is compact, hence for $n\ge1$ exists finite collection $I_n$ of line segments that start at $0$, covering $K $ \ $ R_n$. $R_n = [a,b] \times (\frac{-1}{n},\frac{1}{n})$
While proving compactness, prove that every open set in $(\mathbb{R}^2,d_r)$ containing $[a,b] \times \{0\}$, contains some $R_n$ rectangular.
Is this correct? How to formally write it down?
It takes a fair bit of work to write this out completely, and at least one idea that may not be very obvious. I’ll do most of the harder bits, but I’ll leave a reasonable amount of detail for you to fill in.
You really would be better off working with the actual characterization stated in the theorem. Let’s start with a compact $K\subseteq\Bbb R^2$. In any Hausdorff space every compact set is closed; if you already know that, you can use it here, and if not, you should try to prove it. At any rate, that implies that $K$ is closed in the $d_r$-topology.
Next note that every set that’s open in the usual topology is also open in the $d_r$-topology, so $K$ is also compact in the usual topology on $\Bbb R^2$. This means that $K$ is bounded and hence that there are $a,b\in\Bbb R$ such that $K\subseteq[a,b]\times\Bbb R$. Clearly $K\cap(\Bbb R\times\{0\})\subseteq[a,b]\times\{0\}$, so we need to see what $K\setminus(\Bbb R\times\{0\})$ looks like — that part of $K$ that’s not on the $x$-axis. Specifically, we want to show that there are a set $S=\{s_k:k\in\Bbb Z^+\}\subseteq[a,b]$ and a sequence $\langle t_k:k\in\Bbb Z^+\rangle$ of non-negative real numbers converging to $0$ such that
$$K\setminus(\Bbb R\times\{0\})\subseteq\bigcup_{k\in\Bbb Z^+}\big(\{s_k\}\times[-t_k,t_k]\big)\;.$$
For $n\in\Bbb Z^+$ let $K_n=\left\{\langle x,r\rangle\in K:|r|>\frac1n\right\}$, and let $X_n=\{x\in\Bbb R:(\{x\}\times\Bbb R)\cap K_n\ne\varnothing\}$.
Let $S=\bigcup_{n\in\Bbb Z^+}X_n$; $S$ is the union of countably many finite sets, so $S$ is countable. If $S$ is countably infinite, let $S=\{s_k:k\in\Bbb Z^+\}$. For each $k\in\Bbb Z^+$ there is an $n(k)\in\Bbb Z^+$ such that $s_k\in X_{n(k)}$, and we let $t_k=\frac1{n(k)}$. Otherwise, $S$ is finite, say $S=\{s_1,\ldots,s_m\}$. For $k=1,\ldots,m$ define $t_k$ as before, and for $k>m$ let $t_k=0$.
Prove that $\langle t_k:k\in\Bbb Z^+\rangle$ converges to $0$. Use the fact that each $X_n$ is finite.
Prove that $$K\subseteq\big([a,b]\times\{0\}\big)\cup\bigcup_{k\in\Bbb Z^+}\big(\{s_k\}\times[-t_k,t_k]\big)\;.$$
Once you’ve done all this, you’ll have proved one direction of the theorem.
The other direction is actually the easier direction. Assume that $K$ is closed in the $d_r$-topology and that there are $a,b\in\Bbb R$, a set $S=\{s_k:k\in\Bbb Z^+\}\subseteq[a,b]$, and a sequence $\langle t_k:k\in\Bbb Z^+\rangle$ of non-negative real numbers converging to $0$ such that
$$K\subseteq\big([a,b]\times\{0\}\big)\cup\bigcup_{k\in\Bbb Z^+}\big(\{s_k\}\times[-t_k,t_k]\big)\;;$$
you want to show that $K$ is compact. Compactness is equivalent to limit point compactness in metric spaces, so it suffices to show that $K$ is limit point compact. Suppose, then, that $A\subseteq K$ is infinite.
Show that if $A\cap([a,b]\times\{0\})$ is infinite, then $A$ has a limit point in $[a,b]\times\{0\}$. It may be useful to recognize that $d_r$ and the usual metric are identical on $[a,b]\times\{0\}$.
Show that if $A\cap(\{s_k\}\times[-t_k,t_k])$ is infinite for some $k\in\Bbb Z^+$, then $A$ has a limit point in $\{s_k\}\times[-t_k,t_k]$. It may be useful to recognize that $d_r$ and the usual metric are also identical on $\{s_k\}\times[-t_k,t_k]$.
Prove that if neither of the two previous cases occurs, there is an infinite sequence $\langle k_n:n\in\Bbb Z^+\rangle$ of positive integers such that for each $n\in\Bbb Z^+$ there is a point $$p_n=\langle s_{k_n},y_{k_n}\rangle\in A\cap(\{s_{k_n}\}\times[-t_{k_n},t_{k_n}]\;;$$ then show that the subset $\{p_n:n\in\Bbb Z^+\}$ of $A$ has a limit point in $K$. You’ll want to use the fact that $\{s_{k_n}:n\in\Bbb Z^+\}$ has a limit point in $\Bbb R$, and the fact that $\langle t_{k_n}:n\in\Bbb Z^+\rangle\to 0$.