Let $X$ be a set with at least two elements. Let $\tau, \tau'$ be two topologies on $X$ such that $\tau'\neq\{\emptyset,X\}$. Which of the following condition is necessary for the identity function to be continuous.
$id: (X,\tau)\rightarrow (X,\tau')$
(1) $\tau \subseteq \tau'$
(2) $\tau'\subseteq\tau$
(3) no condition on $\tau$ and $\tau'$
(4) $\tau \cap \tau'=\{\emptyset, X\}$
My Efforts
(1) False
Let $U$ be an open set in $\tau'$, that is not in $\tau$. Inverse image of $U$ is $U$ but the preimage is not open.
(2) True.
Every open set in $\tau'$ is already open in $\tau$ so identity function is continuous.
How should I approve or reject other two option.
What if the topologies are not comparable;?
Given two topologies $\tau,\tau'$ on the same set $X$, the identity function $id:(X,\tau)\to(X,\tau')$ is continuous exactly when $\tau'\subseteq\tau$ for the reason you gave. This forces comparability between the topologies. Thus, the identity function is never continuous when $\tau$ and $\tau'$ are not comparable.
As for why $4)$ is incorrect, think about the continuity condition. We want for every set $U\in\tau'$ that $U\in\tau$. However, if $\tau\cap\tau' = \{\emptyset, X\}$, then this condition fails on every nontrivial open set. It is (in a sense) the 'worst case scenario' for continuity.