How to prove that the induced topology is the coarsest and identification topology is the finest topology that keeps the map continuous?

It's an exercise in definitions:

Recall that given $f:X \to (Y, \tau_Y)$, we define $\tau_f = \{f^{-1}[O]\mid O \in \tau_Y\}$, which is indeed a topology on $X$, from the properties of inverse images you name, among other things.

It's trivial that $f$ is continuous when $X$ is given the induced topology $\tau_f$ wrt $(Y,\tau_Y)$: let $O \in \tau_Y$ be open. By definition of the induced topology $\tau_f$, $f^{-1}[O] \in \tau_f$. So $f$ is continuous as a map $(X,\tau_f) \to (Y, \tau_Y)$.

In fact, if $\tau$ is any topology on $X$ such that $f:(X,\tau) \to (Y, \tau_Y)$ is continuous, when $O \in \tau_f$, we know $O=f^{-1}[O']$ for some $O' \in \tau_Y$, and as $f$ is assumed to be continuous by definition of continuity of $f$, $f^{-1}[O'] \in \tau$, so $O \in \tau$ and $\tau_f \subseteq \tau$. So $\tau_f$ is the coarsest among all topologies that makes $f$ continuous with codomain $(Y, \tau_Y)$.

Now to the identification topology (aka as quotient topology or final topology), when $(X, \tau_X)$ is pre-given and $\tau_i:= \{O \subseteq Y: f^{-1}[O] \in \tau_X\}$ is the identification topology.

Now, if $\tau$ is any topology on $Y$ such that $f:(X, \tau_X) \to (Y, \tau)$ is continuous.

Let $O \in \tau$ arbitrary. Then, as $f$ is continuous, by definition, $f^{-1}[O] \in \tau_X$. But this means that $O$ obeys the defining property for $\tau_i$, so $O \in \tau_i$. And as $O$ is arbitrary, $\tau \subseteq \tau_i$, so the identification topology is the finest (largest) topology among all topologies on $Y$ that make $f$ continuous with domain $(X, \tau_X)$.