If $(X, \tau)$ is a topological space , then for any subset $S$ of $X$ ,$(S, \tau')$ a subspace.where $\tau'$ = [$U \cap S$ : $U$ is an open set in $X$] .
Is the statement true?
I think the statement is true.
If not, can anyone please explain me with counter example why it is not true?
$\tau$ is a collection of subsets of $X.$ In particular, $X$ is in $\tau.$ Thus, if $S$ does not equal $X,$ then there is an element of $\tau$ (namely $X$) that is not a subset of $S.$ Thus, $\tau$ cannot be a topology on $S.$
EDIT: Your new statement is correct. If $S$ is a subset of a topological space $(X, \tau),$ then the subspace topology on $S$ is DEFINED by $$\tau'=\{U \cap S \mid U \in \tau\}.$$ Thus, your statement is true by definition of the subspace topology.