A covering space of a Hausdorff space is Hausdorff 
Let $p:Y\to X$ be a covering space. If $X$ is Hausdorff, so is $Y$.

Hello,
I have a question to this task. I want to show that $Y$ is a Hausdorff space. Hence for $y_1, y_2\in Y$ with $y_1\neq y_2$ have disjoint neighbourhoods.
I want to prove this:
Let $p:Y\to X$ be a covering space.
$y_1, y_2\in Y$, with $y_1\neq y_2$.
$x_1:=p(y_1)$ and $x_2:=p(y_2)$. Then exist open sets $x_1\in U_1\subseteq X$ and $x_2\in U_2\subseteq X$, such that $p^{-1}(U_i)$ with $i=1,2$ is the disjoint union of open sets.
Hence there are open sets $y_1\in V_1\subseteq p^{-1}(U_1)$ and $y_2\in V_2\subseteq p^{-1}(U_2)$.
Claim: $V_1\cap V_2=\emptyset$.
I do not think, that this works...
Is this approach any good? 
Do you have tips on how I can do better?
Thanks in advance.

Hint: 
First case $x_1\neq x_2$, since $X$ is separated, there exists $U_1,U_2$ open neighborhoods of respectively $x_1$ and $x_2$ such that $U_1\cap U_2$ is empty. $p^{-1}(U_1)\cap p^{-1}(U_2)$ is empty and $y_1\in p^{-1}(U_1), y_2\in p^{-1}(U_2)$.
Second case $x_1=x_2=x$, since $p$ is a covering, there exists a neighborhoo $U$ of $x$ such that  for every $y\in p^{-1}(x)$, there exists a neighborhood $V_y$ such that the restriction of $p$ to $V_y$ is an homeomorphism $p_{\mid V_y}:V_y\rightarrow U$, and $V_{y_1}\cap V_{y_2}$ is empty by definition.
