If a subset of $\mathbb{R}$ is closed and bounded with respect to a metric equivalent to the Euclidean metric, must it be compact? 
Two different metrics $d$ and $\hat d$  in a space $X$ are said to be equivalent iff the topologies generated by them are the same, in other words $U\subseteq X$ is $d$-open iff it is $\hat d$-open.
So by definition $(c)$ is correct, am I right?
 A: No. Even if $d$ and $\hat{d}$ induce the same topology (in which case a set is $d$-closed $\iff$ it is $\hat{d}$-closed), it need not be the case that $\hat{d}$-bounded $\implies$ $d$-bounded, so one cannot conclude from the Heine-Borel theorem ($d$-closed and $d$-bounded $\implies$ $d$-compact for subsets of $\mathbb{R}^n$) that $\hat{d}$-closed and $\hat{d}$-bounded $\implies$ $\hat{d}$-compact for subsets of $\mathbb{R}^n$.
BenjaLim has given the example $\hat{d}(x,y)=\min\{d(x,y),1\}$. The other standard example is $$\hat{d}(x,y)=\frac{d(x,y)}{1+d(x,y)}.$$
One can show that for any metric space $(X,d)$, both of these definitions of $\hat{d}$ induce the same topology as $d$, but that $X$ is bounded under these $\hat{d}$ even if it was not under $d$. 
Remember that boundedness is strictly a property that comes from a metric, i.e. boundedness is not a topological property.
A: (c) is false, as an example take $\hat{d}(x,y) = \min\{d(x,y),1\}$. Then you can check that $\hat{d}$ and $d$ generate the same topology. Now $\Bbb{R}$ is closed and bounded with respect to the $\hat{d}$ metric, but clearly it cannot be compact: Any open cover of $\Bbb{R}$ in the $\hat{d}$ metric is also an open cover  for $\Bbb{R}$ in the $d$ metric which has no finite subcover.
