If $f:X\to Y$ is a one-to-one continuous mapping, and $Y$ is a $T_{2}$-space, then does $X$ too have to be a $T_{2}$-space? Motivation:
For any two points $y_{1},y_{2}\in Y$, there are disjoint open sets containing $y_{1}$ and $y_{2}$ separately, as $Y$ is a $T_{2}$-space. 
Say $f(x_{1})=y_{1}$ and $f(x_{2})=y_{2}$. Then, taking the inverses of the disjoint open sets, we get disjoint open sets containing $x_{1}$ and $x_{2}$ separately. 
Why the inverses of disjoint open sets are also disjoint is if there was one point in common between two open sets in $X$ and not in $Y$, then that point would mapped to two different points in $Y$, one in each disjoint set, which is impossible. 
Remember $f$ is one-to-one. Hence, as there are disjoint open sets for every $y_{i}$ and $y_{j}$, there are correponding open sets for every $x_{i}$ and $x_{j}$. Doesn't this make $X$ a Hausdorff space too? 
 A: You  are right. However we can conclude that 

If $f:X\to Y$ is a one-to-one continuous mapping, and $Y$ is a $T_{3}$-space, then $X$ don't have to be $T_3$.

Example 1: Consider the space $X$ with Minimal Hausdorff Topology. It is $T_2$, non $T_3$. However it is submetrizable, i.e., there exists a one-to-one continuous mapping from $X$ to a metrizable space. Note that every metrizable space is $T_3$.
$X$ is submetrizable.

Prove: $X$ is countable, then $X$ is separable and has a zeroset diagonal. Every separable space with a zeroset diagonal is submetrizable.

A: You are right about the situation when $f: X \to Y$ is continuous, 1-1 and onto, and $Y$ is Hausdorff then so is $X$. The proof you sketched is correct: the images of distinct points are distinct and they have disjoint neighbourhoods that we can pull back.
In the comments you asked about openness. It is not true that $f$ needs to be an open map. A standard example: $X = [0,2\pi)$, $Y= S^1 \subset \mathbb{R}^2$, the unit circle, and $f(x) = (\cos(x),\sin(x))$. This map is not open. E.g. consider $f[[0,\pi)]$ and note that $[0,\pi)$ is open in $[0,2\pi)$.
But if $X$ is compact as well, then the map $f$ will be closed and so $f$ is a homeomorphism (and thus open as well). 
