$A = \left\{ (1,x) \in \mathbb{R}^2 : x \in [2,4] \right\} \subseteq \mathbb{R}^2$ is bounded and closed but not compact? Is it true that set $A = \left\{ (1,x) \in \mathbb{R}^2 : x \in [2,4] \right\} \subseteq \mathbb{R}^2$ is bounded and closed but is not compact. We consider space $(\mathbb{R}^2, d_C)$ where $$d_C(x,y) =  \begin{cases} d_E(x,y) \quad \text{if x, y, 0 in one line} \\ d_E(x,0)+d_E(0,y) \quad \text{in any other case} \end{cases}$$
Where of course $d_E$ is euclidean metric.
 A: Yes, it’s true. In fact, $A$ is a closed, discrete set in $\Bbb R^2$ with the metric $d_C$, and an infinite, discrete set is never compact.
For each $a\in[2,4]$ let 
$$U_a=\left\{\langle x,ax\rangle:\frac12<x<\frac32\right\}\;;$$
$U_a$ is an open interval on the line through $\langle 0,0\rangle$ and $\langle 1,a\rangle\in A$, so it’s an open set in the space; in fact, it’s the open $d_C$-ball of radius $\frac{\sqrt{a^2+1}}2$ centred at $\langle 1,a\rangle$. $U_a\cap A=\langle 1,a\rangle$ for each $a\in[2,4]$, so $A$ is discrete. In fact, the open sets $U_a$ for $a\in[2,4]$ are even pairwise disjoint. Thus, $\{U_a:a\in[2,4]\}$ is an open cover of $A$ that has no finite subcover. (It even has no proper subcover: every one of the sets $U_a$ is actually needed in order to cover $A$.)
The point of $A$ furthest from the origin is $\langle 1,4\rangle$, and $d_C(\langle 0,0\rangle,\langle 1,4\rangle)=\sqrt{17}$, so $A$ lies within a ball of finite radius centred at the origin and is therefore bounded. I’ll leave it to you to verify that $A$ is closed in this topology; the ideas that I used to construct the sets $U_a$ should help, if you don’t already see how.
A: Yes, that is right. Note that each point $a\in A$ is isolated, because if you choose $\epsilon<d_E(a,0)$, then this ball does not contain any other point $b\in A$, as the distance $d_C(a,b)$ would be larger than $d_E(a,0)$. That means that $A$ is discrete. 
$A$ is bounded since the distance from each point in $A$ to $0$ is just the Euclidean distance.
Finally, $A$ is closed:


*

*$0$ has its Euclidean distance from $A$ which is clearly positive.

*Each point $y=(y_1,y_2)\ne0$ not in $A$ has a $d_E(y,0)$-ball not containing any point on a different line through the origin.

*Such a point also has a positive distance from $\left(1,\frac{y_2}{y_1}\right)$, the only possible point in $A$ which can be on the same line through $0$, namely $(1-y_1)\sqrt{1+\left(\frac{y_2}{y_1}\right)^2}$

