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.

You need to break your proof up into two parts. 
Suppose $x_1 \neq x_2$. We can choose $U_1$ and $U_2$ to be disjoint, and so $p^{-1}(U_1)$ and $p^{-1}(U_2)$ will be disjoint (if they had a point $z$ in common, $p(z) \in U_1\cap U_2 = \emptyset$ which is impossible). Now you can take $V_1 = p^{-1}(U)$ and $V_2 = p^{-1}(U_2)$. 
Suppose $x_1 = x_2$; call this point $x$. Let $U$ be an evenly covered open neighbourhood of $x \in X$, then $p^{-1}(U) \cong \bigsqcup_{\alpha} U_{\alpha}$ where $p|_{U_{\alpha}} : U_{\alpha} \to U$ is a homeomorphism. Suppose $y_1 \in U_{\beta_1}$ and $y_2 \in U_{\beta_2}$, then it follows that $\beta_1 \neq \beta_2$. To see this, suppose $\beta_1 = \beta_2$; call this value $\beta$. As $p|_{U_{\beta}}$ is a homeomorphism and $p|_{U_{\beta}}(y_1) = p|_{U_{\beta}}(y_2)$, we have $y_1 = y_2$; this is a contradiction as $y_1$ and $y_2$ were assumed to be distinct. As $\beta_1 \neq \beta_2$, we can take $V_1 = U_{\beta_1}$ and $V_2 = U_{\beta_2}$.
