Pre-image of Hausdorff space is Hausdorff
Let $x_1,x_2 \in X$ such that $x_1 \neq x_2$
Then $f(x_1),f(x_2) \in Y$ and $f(x_1) \neq f(x_2)$ because $f$ is injective.
Then there exist two disjoint open sets $U_1,U_2 \in \mathcal{T_Y}$ such that $f(x_1) \in U_1$ and $f(x_2) \in U_2$
Thus $$x_1 \in f^{-1}(U_1)$$ $$x_2 \in f^{-1}(U_2)$$ where $ f^{-1}(U_1),f^{-1}(U_2)$ are open in $X$ because $f$ is continuous and $U_1,U_2$ are open in $Y$.
Put $V_1=f^{-1}(U_1)$ and $V_2=f^{-1}(U_2)$ and you have: $$V_1 \cap V_2 =f^{-1}(U_1) \cap f^{-1}(U_2)=f^{-1}(U_1 \cap U_2)=f^{-1}(\emptyset)=\emptyset$$
Therefore $X$ is Hausdorf.
There's a jump in your proof: you write that $x$ and $y$ have disjoint neighborhoods $U\in\tau_x$ and $V\in\tau_y$ (note that you wrote $U\subset\tau_x$ and $V\subset\tau_y$; this is wrong), and, in the next sentence, you write that the pre-images of $U$ and $V$ are open sets. Why?
There are two ways of solving this:
- Use the fact that the pre-image of a neighborhood by a continuous function is again a neighborhood.
- You can assume without loss of generality that $U$ and $V$ are open sets.
Take distinct $x,y \in X$ then $f(x), f(y)$ are distinct points in $Y$ (by injectivity) and so we have disjoint open sets $U$ and $V$ containing $f(x)$ and $f(y)$, respectively. Now their preimages $f^{-1}(U)$ and $f^{-1}(V)$ are open (by continuity), disjoint and contain $x$ and $y$.