Let $f: (X, \mathcal{T}_2) \rightarrow (X, \mathcal{T}_1)$ be a bijection between two topological spaces. Prove that $f$ is continuous if and only if $\mathcal{T}_1 \subset \mathcal{T}_2$
Here is my proof for the backward direction:
Suppose $\mathcal{T}_1 \subset \mathcal{T}_2$ but $f$ is not continuous. Then there is a $U \in \mathcal{T}_1$ such that $f^{-1}(U) \notin \mathcal{T}_2$. Then since $f$ is bijective, $U \in \mathcal{T}_1 \subset f(\mathcal{T}_2)$, which is a contradiction.
For the forward direction, suppose $f$ is continuous but $\mathcal{T}_1 \not\subset \mathcal{T}_2$. Then there is a $U \in \mathcal{T}_1$ such that $U \notin \mathcal{T}_2$. By the continuity of $f$, $f^{-1}(U) \in \mathcal{T}_2$, and here is where I'm not sure how to proceed.
Unfortunately neither implication is true. Let
$$X=\{1,2\}$$ $$\mathcal{T}_1=\{\emptyset, X, \{1\}\}$$ $$\mathcal{T}_2=\{\emptyset, X, \{2\}\}$$
Now let $f:X\to X$ be given by $f(1)=2$ and $f(2)=1$. You can easily check that $f$ is continuous (in fact it is a homeomorphism) but neither $\mathcal{T}_1\subseteq \mathcal{T}_2$ nor $\mathcal{T}_2\subseteq \mathcal{T}_1$.
On the other hand let
$$X=\{1,2\}$$ $$\mathcal{T}_1=\{\emptyset, X, \{1\}\}$$ $$\mathcal{T}_2=\{\emptyset, X, \{1\}\}$$
This time $\mathcal{T}_1\subseteq \mathcal{T}_2$ (they are even equal). Again consider $f(1)=2$ and $f(2)=1$ and note that it is not continuous.
The statement is not true for general bijection $f$. It is true however for the identity map $f(x)=x$.