is $(0,1)$ compact in indiscrete topology and discrete topolgy on $\mathbb R$ is $(0,1)$ compact in indiscrete topology and discrete topolgy on $\mathbb R$? i think $(0,1)$ is compact in discrete topology on $\mathbb R$ because it closed and bounded. but the same set is not compact in indiscrete topology on $\mathbb R$ because it is not closed (because in indiscrete topolgy on $\mathbb R$ the closed sets is only $\phi$ and $\mathbb R$). but my teacher say wrong answer :( why ?

In the discrete topology, one point sets are open. So you can take the cover by those sets. Removing just one element of the cover breaks the cover. In fact no infinite set in the discrete topology is compact.

As for the indiscrete topology, every set is compact because there is only one possible open cover, namely the space itself. Since that cover is finite already, every set is compact.