Ball around a set equals union of balls around each of its points I have this problem:
Let $(X,d)$ be a metric space and let $A \subseteq X$. Define for any $ \epsilon > 0$:
    $$B_d(A,\epsilon) = \left\{ x \in X \ : \ d(x,A) < \epsilon \right\}.$$
    Show that $B_d(A,\epsilon) = \bigcup_{x \in A} B_d(x,\epsilon)$.
I was trying to show the double inclusion, but I am not sure about anything. I know that $d(x,A)$ is defined as:
$$d(x,A) = \inf \left\{d(x,a) | a \in A\right\}.$$
My attempt went as follows:
\begin{align*}
y \in B_d(A,\epsilon) &\Rightarrow d(y,A) < \epsilon\\
                 & \Rightarrow \inf \left\{d(x,a) | a \in A\right\} < \epsilon\\
& \Rightarrow \exists z \in A (d(y,z) < \epsilon)\\
&\Rightarrow \exists z \in A (y \in B_d(z,\epsilon))\\
&\Rightarrow y \in \cup_{z \in A} B_d(z,\epsilon).
\end{align*}
I doubt this is correct. And I don't know about the other inclusion. Does anybody have a very rigorous proof of this fact?

Filling the minor gap in the forward inclusion:
Take $p \in B_d(A,\varepsilon)$. So $d(p,A) < \varepsilon$. Suppose that for all $x \in A$ we would have that $\varepsilon \le d(x,p)$.
This means that $\varepsilon$ is an lowerbound for the set $\{d(x,p): x \in A\}$ and $d(p,A)= \inf \{d(x,p): x \in A\}$ is the maximum of the set of lowerbounds and we'd have $d(p,A) \ge \varepsilon$ which is a contradiction.
So for some $x \in A$, $d(x,p) < \varepsilon$ or equivalently $p \in B_d(x,\varepsilon)$ for that $x$, and hence
$$B_d(A,\varepsilon)\subseteq \bigcup_{x \in A} B_d(x,\varepsilon)$$
For the reverse, if for some $x \in A$ we have $p \in B_d(x,\varepsilon)$ we know that $d(x,p) < \varepsilon$ and so $d(p,A) \le d(x,p)$ (the infimum of a set is a lowerbound of that set) and so $d(p,A) < \varepsilon$ or $p \in B_d(A,\varepsilon)$ and so
$$B_d(A,\varepsilon)\supseteq \bigcup_{x \in A} B_d(x,\varepsilon)$$
and equality ensues.
