$A = \left\{ (1,x) \in \mathbb{R}^2 : x \in [2,4] \right\} \subseteq \mathbb{R}^2$ 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.

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 $(1-y_1)\sqrt{1+\left(\frac{y_2}{y_1}\right)^2}$

