Show that $d(b, X) = d(b, \overline{X})$ I need to prove that, in a vector space $V$, given $b \in V$, we have $d(b, X) = d(b, \overline{X})$, where $X = B(a;r) = \{x \in V \mid ||x - a|| < r\}$, and $\overline{X} = B[a;r] = \{x \in V \mid ||x - a|| \leq r\}$.
I supposed that $m = d(b, \overline{X})$ and proved that $m$ is a lower bound for the set $\{||b-x|| \mid x \in X\}$. Now I need to construct some vector $x \in X$ such that, for $\epsilon > m$, we have $||b-x|| < \epsilon$, i.e., $\epsilon$ can't be an lower bound for this set of distances in $X$.
I'm struggling to construct this $x$ using the fact that $V$ is a vector space. Any tips are helpful
 A: We can eventually show $d(b,X)=d(b,\overline{X})$ by considering the definitions of infimum separately and appealing to equality as a double inequality (standard trick).

To help us think, let's rename:
$$\color{red}{S_{o}}:= \{d(b,x)\text{ }|\text{ }x\in X\}\text{ }\text{ }\text{ , }\text{ }\text{ }\color{red}{S_{c}}:= \{d(b,x)\text{ }|\text{ }x\in \overline{X}\}$$
$$\color{red}{\iota_{o}} := \inf S_{o}\text{ }\text{ }\text{ and }\text{ }\text{ }\color{red}{\iota_{c}}:=\inf S_{c}.$$

The (indirect) definition of infimum as greatest lower bound gives:
$$(1a)\text{ }\text{ }\text{ }\text{ }\forall x\in X:\text{ }\iota_o\leq d(b,x)\text{ }\text{ }\Bigg|\text{ }\text{ }(1b)\text{ }\text{ }\text{ }\text{ }\forall x\in \overline{X}:\text{ }\iota_c\leq d(b,x)$$
$$(2a)\text{ }\text{ }\text{ }\text{ }\forall LB_{S_o}:\text{ }LB\leq\iota_o\text{ }\text{ }\Bigg|\text{ }\text{ }(2b)\text{ }\text{ }\text{ }\text{ }\forall LB_{S_c}:\text{ }LB\leq\iota_c.$$
Excuse the shorthand, $LB_{S_o}$ here means "lower bound for $S_o$" etc.

$\underline{Proof:}$
$\boldsymbol{(\geq):}$
The fact that $X\subseteq \overline{X}$ allows us to restrict $(1b)$ to get:
$$\forall x\in X:\text{ }\iota_c\leq d(b,x),$$
but this says $\iota_c$ is a $LB_{S_o}$, so by $(2a)$, it follows that:
$$\iota_c\leq \iota_o.$$
$\boldsymbol{(\leq):}$
By $(1a)$ we have:
$$\forall x\in X\text{ }\big(\supseteq\overline{X}\backslash \partial X\big): \iota_o\leq d(b,x).$$
If we can show:
$$\color{blue}{\forall x\in \partial X: \iota_o\leq d(b,x)}$$
then $\iota_o$ is a $LB_{S_c}$ and hence we would have by $(2b)$ that:
$$\iota_o\leq \iota_c.$$
Combining these results gives $(\iota_o = \iota_c)$. In other words: $d(b,X) = d(b,\overline{X}).\text{ }\text{ }\square$ 
$\text{ }\text{ }$ Let's show the blue statement then:
In our case, for all $x\in \partial X$, there exists a sequence $\{x_n\}_{n\in\mathbb{N}}\subseteq X$ such that $x_n\to x$. This is easy to show using the center of the ball, '$a$', and parameterizing a discrete line to $x$.
Since the metric, $d$, is continuous in both variables, we get in particular that $d$ preserves limits of sequences in either variable. So:
$$d(b,x) = d(b,\lim\limits_{n\to\infty}x_n) = \lim\limits_{n\to\infty}d(b,x_n).$$
Now, by containment of the sequence in $X$ and $(1a),$ we have a sequence of inequalities:
$$\forall n\in \mathbb{N}:\text{ }\iota_o\leq d(b,x_n).$$
Thus, in the limit, we have:
$$\iota_o \leq d(b,x).$$
Arbitrariness of $x\in \partial X$ completes the proof.$\text{ }\text{ }\blacksquare$
