I've found this exercise.
Let $X,Y$ be two topological spaces and $f$ an homeomorphism between $X$ and $Y$. Show that $X$ is $T_1$ iff $Y$ is $T_1$.
Let $(Y,\tau')$ be a $T_1$ space. So $Y\setminus y$ is an open set.
Since $f$ is an homeomorphism, $f^{-1}$ is a continuous function, then $f^{-1}(Y \setminus y)=X \setminus f^{-1}(y) \in \tau$, and since $f^{-1}(y)$ is a point, by the arbitrary of $y$ follows that $X$ is a $T_1$ space.
Let $(X,\tau)$ be $T_1$.
So $X \setminus x$ is open in $X$. Since $f$ is an homeomorphism, $f$ is an open map,so $f(X \setminus x)=Y \setminus f(x) \in \tau'$, and follows that $Y$ is $T_1$
You start your proof mentioning $y$, without saying what it is. You should have added “Let $y$ be an arbitrary point of $Y$.” or something similar. But it would be a better option to start with a point $x\in X$: since $Y$ is $T_1$, $Y\setminus\{f(x)\}$ is open and, since $X\setminus\{x\}=f^{-1}\bigl(Y\setminus\{f(x)\}\bigr)$, $X\setminus\{x\}$ is open.
For the proof in the opposite direction, you could have just said that the same argument applies with the roles of $X$ and $Y$ reversed, since $f^{-1}$ is a homeomorphism too.