Exercise: Let $\tau_1$ and $\tau_2$ be two topological spaces on a set $X$. Then $\tau_1$ is said to be a finer topology than $\tau_2$ if $\tau_1\supseteq\tau_2$.
Prove the identity function $f:(X,\tau_1)\to (X,\tau_2)$ is continuous if and only if $\tau_1$ is a finer topology than $\tau_2$.
Attempted proof:
Let $A\in\tau_1$. If $f$ is continuous then $\exists B\in\tau_2$ such $f(A)\subset B\implies A\subset B$ for any $A\in\tau_1$ such that all open sets of $\tau_1$ are open sets of $\tau_2$ hence $\tau_1\supseteq \tau_2$.
If $\tau_1$ is a finer topology than $\tau_2$, for any $B\in\tau_2\implies B\in\tau_1$ which implies that $f^{-1}(B)=B$, then $B\in\tau_1$ which proves $f$ to be continuous.
Questions:
Is my proof right? If not. What is wrong? What are alternatives?
Thanks in advance!
When working with continuity, you always have to use the pre-image, I don't understand what you are trying to prove for $\implies$.
The direction $\Longleftarrow $ is correct as you have shown it.
For $\implies$ I would have just done the following: for any open set $B$ in $\tau_2$ as $f$ is continuous, $f^{-1}(B)\in \tau_1$ (i.e. is open) by definition. But as $f=id$, $f^{-1}(B)=B\in\tau_1$ So $\tau_2\subseteq \tau_1$, as wanted.