Consider the space $X =[0,3]$ (endowed with subspace topology). Consider the subspace $(1,2)$. Since $(1,2)$ is not compact, there exists an open cover $\mathscr G=\{G_\alpha\}_{\alpha\in \Lambda}$ of $(1,2)$ such that it has no finite subcover. Since $(1,2)$ is open in $X$, it's open sets correspond to those of $X$.
Now, consider an open cover $\mathscr H=\{H_\beta\}_{\beta\in \Gamma}$ of $X$ such that $\mathscr G\subset\mathscr H$.
Since $X$ is compact, $\mathscr H$ has a finite subcover. But, $\mathscr G$ doesn't have a finite subcover, and it being a subset of $\mathscr H$, should have one, thus a contradiction.
Now, this is definitely an incorrect argument, since the properties of both the spaces are well established. Where exactly am I going wrong?
Even though $\mathscr{G} \subset \mathscr{H}$, the finitely many of the sets in $\mathscr{H}$ that cover $[ 0, 3]$ and hence also $(1, 2)$ may well include some that do not belong to $\mathscr{G}$.
For example, let $\mathscr{G}$ consist of all the open intervals of the form $\left( 1, 2 - \frac1n \right)$ for $n = 1, 2, 3, \ldots$. This $\mathscr{G}$ of course covers $(1, 2)$ but no finite subcollection of $\mathscr{G}$ does.
Now let $\mathscr{H}$ be any open cover of $[0, 3]$ such that $\mathscr{H} \supset \mathscr{G}$. This $\mathscr{H}$ does have sets other than those in $\mathscr{G}$, and since $$ \bigcup_{n = 1}^\infty \left( 1, 2 - \frac1n \right) = (1, 2), $$ any finite subcollection of $\mathscr{H}$ that could cover $[0, 3]$ --- and such a finite subcollection we know does exist --- necessarily must include one or more sets that are not in $\mathscr{G}$.