Proof verification: All metric spaces gave the "Hausdorff property". I am currently studying an introductory book to complex analysis. Within the first chaper the complex numbers are presented in various ways aswell as their structure regarding to different points of view $($e.g. as a vector space, in polar form, as group, etc.$)$.
However, I am having a hard time understanding the concept of topology which is also used there. Especially I am not quite sure whether I have fully understood the idea behind the proofs. Therefore $-$ and moreover in order to pratice proof writing $-$ I picked out a theorem and tried to prove it all by myself. Nevertheless I am not sure if I have managed to work out a sufficient proof or if I missed a crucial point.

Currently I am dealing with topological spaces, metric spaces, and their properties aswell as their relations. Firstly define a topological spaces as a metric space:

All metric spaces are topological spaces
  Every metric space $X$ is a topological space, where an open set is defined as the following: a set $M\subset X$ is called "open" if there exists for every $x\in X$ an $\varepsilon>0$, such that the "$\varepsilon$-environment" $$U_{\varepsilon}(x_0):=\{x\in X:d(x,x_0)<\varepsilon\}$$ is completely within $M$.  

Here $d(x,y)$ denotes the specific metric of the metric space. Secondly there is a so-called "Hausdorff property" stated as:

"Hausdorff property"
  A topological space $X$ is a Hausdorff-space, if there exist for every distinct pair of points $x\ne y$ enviroments $U=U(x)$ and $V=V(y)$ such that $U\cap V=\emptyset$.

My aim is to prove that every metric space satisfies the "Hausdorff-property".
Proof
Assume there exists an point $y$ such that $y\in U_{\varepsilon}(x)$ and $y\in V_{\varepsilon'}(x')$ where $x\ne x'$. By definition the enviroments  $U_{\varepsilon}$ and $V_{\varepsilon'}$ are defined as
$$\begin{align}
U_{\varepsilon}(x)&=\{y\in X:d(y,x)<\varepsilon\}\\
V_{\varepsilon'}(x')&=\{y\in X:d(y,x')<\varepsilon'\}
\end{align}$$
From $x\ne x'$ we can conclude that $d(x,x')>0$. Furthermore the triangle inequality states that $d(x,x')\le d(x,y)+d(x',y)$. Consequently this leads to
$$\begin{align}
d(x,x')&\le d(x,y)+d(x',y)\\
0<d(x,x')&\le d(x,y)+d(x',y)\\
0&< d(x,y)+d(x',y)\\
0&< d(x,y)+d(x',y)<\varepsilon+\varepsilon'
\end{align}$$
Since $\varepsilon,\varepsilon'$ are choosen to be small numbers $($?$)$ this further yields to the contradiction
$$0<d(x,y)+d(x',y)<0$$
From where can conclude that $d(x,y)=d(x',y)=0$. Since $d(x,y)$ only equals $0$ iff $x=y$ it follows that $x'=x=y$ which contradicts our assumption.$\small\square$ 

As marked within my proof I am not sure whether my argumentation about the $\varepsilon,\varepsilon'$ is valid. Anyway I do not know how to finish the proof otherwise.

Is this proof legitimate or not? In the case this is right, what can be improved? In the case it is not, where have I gone wrong? I would appreciate if you could follow my own attempt, i.e. by using the metric of the space in order to prove the "Hausdorff-property". Anyway if this is an impractical approach I am open for different ways of showing the given thereom.

Thanks in advance!
 A: The main issue with your proof is that you haven't actually said what $\varepsilon$ and $\varepsilon'$ are before they're used, and then later on you say they're "small numbers", but again this is too vague.
You then conclude that $d(x,y) + d(x',y) < 0$, but this conclusion is erroneous for two reasons:


*

*$\varepsilon$ and $\varepsilon'$ are (or, rather, should be) fixed from the start, and so you can't jump from $d(x,y) + d(x',y) < \varepsilon + \varepsilon'$ to this conclusion.

*Even if you had proved that $d(x,y) + d(x',y) < \varepsilon + \varepsilon'$ for all $\varepsilon,\varepsilon' > 0$ (which you can't, since they're supposed to be fixed), the conclusion that you'd actually end up drawing is that $d(x,y) + d(x',y) \le 0$. But this is a moot point, as I mentioned.


Here's how I'd suggest to fix your proof. Take $x,x' \in X$ with $x \ne x'$. Then $d(x,x') > 0$, and so you can let $\varepsilon = \dfrac{d(x,x')}{2}$, which is positive (and is now fixed!).
Use a proof strategy similar to what you already tried, in order to prove that $U_{\varepsilon}(x)$ and $U_{\varepsilon}(x')$ are disjoint open neighbourhoods of $x$ and $x'$, respectively.
A: You wrote that $\varepsilon,\varepsilon'$ were chosen to be small numbers, but you did not tell us which choice was that. Therefore, you proved nothing.
Simply take $\displaystyle\varepsilon=\varepsilon'=\frac{d(x,x')}2$. That will do: $U_\varepsilon(x)\cap U_{\varepsilon'}(x')=\emptyset.$
