Show when $x\neq y$ then in a Metric space $d(x,y):= \epsilon > 0$

Question

Let $X$ be a Metric space and $x\neq y$. Show that $d(x,y):= \epsilon > 0$ and I am given as a hint to show that that $B_{ \frac{\epsilon}{2}}(x)\cap B_{ \frac{\epsilon}{2}}(y)=\varnothing$

My idea:

Assume that $B_{ \frac{\epsilon}{2}}(x)\cap B_{ \frac{\epsilon}{2}}(y)\neq\varnothing \Rightarrow z \in B_{ \frac{\epsilon}{2}}(x)\cap B_{ \frac{\epsilon}{2}}(y)\Rightarrow \epsilon :=d(x,y)\leq d(x,z)+d(z,y)<\epsilon$ which is a contradiction. Thus $B_{ \frac{\epsilon}{2}}(x)\cap B_{ \frac{\epsilon}{2}}(y)=\varnothing$

I have shown the hint, but how does this prove that $\epsilon > 0$?

Answer 1

In the definition of a metric, there is the clause (Or a variation of it):

$$\forall x,y \in X: d(x,y)=0 \iff x=y\tag{1}$$

Plus the fact that $d$ is defined a function from $X \times X \to \mathbb{R}^+$ (So that all values are $\ge 0$ implicitly) or as an a function to $\mathbb{R}$ with an extra explicit

$$\forall x,y \in X: d(x,y) \ge 0\tag{2}$$

The combination of these facts gives that if $x \neq y$, then $d(x,y) \ge 0$ by (2), or its implicit version, but $d(x,y)=0$ is ruled out (1), as $d(x,y)=0$ would imply $x=y$ while we know that $x \neq y$. And if a real number is $\ge 0$ but not $0$ by definition it's $>0$.