Using arbitrariness of $\varepsilon$ in proofs (metric spaces etc) Let $(X, d)$ be a metric space. Let $A \subseteq X$ and fix $x \in X$, define $\rho(x, A) = \inf \{d(x,a) : a \in A \}$. 
In trying to prove $\overline{A} = \{x \in X : \rho(x, A) = 0\}$, if $x \in \overline{A}$ then for all $\varepsilon>0$, $B_{\varepsilon}(x) \cap A \neq \emptyset$. So let $y_{\varepsilon} \in B_{\varepsilon}(x) \cap A$. By the definition of infimum, we have
\begin{align*}
\rho(x, A) \le d(x, y_{\varepsilon}) < \varepsilon.
\end{align*}
Then since $\varepsilon$ was arbitrary, we have $\rho(x, A) = 0$.
What I don't understand is the bolded part, why is it that just because $\varepsilon$ is arbitrary allows us to conclude $\rho(x, A) =0$? What is the fundamental reason behind this? I've seen this kind of argument using the arbitrariness of $\varepsilon$ used elsewhere in really simple arguments, for example:
For all $\varepsilon>0$, assume that $x'+y' > u_1 + u_2 - \varepsilon$ for some real numbers $x', y', u_1, u_2$. Then since $\varepsilon$ was arbitrary, we have $x'+y' \ge u_1 + u_2$. Why? I think I'm missing something really obvious.

Assume that $\rho(x,A)\neq 0$, then $\varepsilon:=\rho(x,A)/2>0$ and by assumption, one has:
$$\rho(x,A)<\varepsilon=\rho(x,A)/2.$$
Which leads to $1<1/2$, a contradiction.
