help me find counterexample for two comparable topologies on a set X, compactness of the bigger one (bigger in the sense of containment) implies the compactness of the smaller one, can anyone help me find example of two comparable topologies where compactness is there in the smaller topology and not in the bigger one?
( smaller and bigger according to containment)
 A: Let $\langle X,\tau\rangle$ be any compact Hausdorff space, and let $\tau'$ be any topology strictly finer than $\tau$; then $\langle X,\tau'\rangle$ is not compact.

Proof. The identity map $\mathrm{id}_X:X\to X:x\mapsto x$ is continuous as a map from $\langle X,\tau'\rangle$ to $\langle X,\tau\rangle$, since for each $U\in\tau$ we have $\mathrm{id}_X^{-1}[U]=U\in\tau'$. Since $\tau'\supsetneqq\tau$, there is a $K\subseteq X$ such that $X\setminus K\in\tau'\setminus\tau$, i.e., such that $K$ is $\tau'$-closed but not $\tau$-closed. Suppose that $\langle X,\tau'\rangle$ is compact. Then the $\tau'$-closed set $K$ is compact in $\langle X,\tau'\rangle$, so its continuous image $\mathrm{id}_X[K]=K$ is compact in $\langle X,\tau\rangle$, contradicting the fact that every compact set in a Hausdorff space is closed. $\dashv$

This result makes it easy to find examples. Taking $\tau'$ to be the discrete topology on $X$ is of course one way that always works, so long as $X$ is infinite. A couple of more interesting examples are the Sorgenfrey (or lower-limit) topology on $[0,1]$ and the Michael line topology on $[0,1]$. The former has as a base all intervals of the form $[a,b)$ with $0\le a<b\le 1$ together with the singleton $\{1\}$. The latter is obtained by isolating each irrational number in $[0,1]$, so if $\tau$ is the usual topology on $[0,1]$, the Michael line topology has as a base the family $\tau\cup\big\{\{x\}:x\in[0,1]\setminus\Bbb Q\big\}$. Both of these topologies are quite nice; you’ll find some of their nice properties listed at the links.
A: "Finer" and "coarser" are words I'm accustomed to in this context.  $T$ is finer than $S$, and $S$ is coarser than $T$, iff $S\subseteq T$.  If $X$ is compact with the finer topology, then $X$ is compact with the coarser one (so it seems to me you had it backwards).  Take the sphere with the usual topology and also with the discrete topology; it's compact in the usual one and not in the discrete one.  That's a bit extreme.  Probably there are subtler and more interesting examples.
