Complementing the excellent answers by @bof, here is the final case:
Theorem. Let $X$ be a normed space. Then the following are equivalent:
i) $X$ is complete,
ii) For every totally ordered family $\mathscr B$ of open balls in $X$ with uniformly bounded radii, the union of the members
of $\mathscr B$ is an open ball.
Proof (i) $\Rightarrow$ (ii) was already proven by @bof in their accepted answer.
(ii) $\Rightarrow$ (i): Assuming (ii), and arguing by contradiction, suppose that $X$ is not complete. Denoting by
$\tilde X$ the completion of $X$, let $a$ be a point of $\tilde X$ which is not in $X$. Choose a sequence $\{c_n\}_{n=1}^\infty $
in $X$ such that $\|c_n-a\|<1/2^{n+1}$, so that $c_n\to a$, as $n\to\infty$.
It follows that
$$
\|c_n-c_{n+1}\| \leq
\|c_n-a\| + \|a-c_{n+1}\| \leq
{1\over 2^{n+1}} + {1\over 2^{n+2}} < {1\over 2^n}.
\tag 1
$$
Define
$$
r_n = \sum_{k=0}^{n-1} {1\over 2^k}, \quad \text{and} \quad r = \sum_{k=0}^\infty {1\over 2^k}
$$
(of course we could spell out the explicit values of $r_n$ and $r$, but the above expressions will turn out to be more
convenient for us) and observe that
$$
r_{n+1} = r_n + {1\over 2^n},
\tag 2
$$
and that $r_n\to r$, as $n\to\infty $.
Setting
$$
\tilde B_n=B^{\tilde X}_{r_n}(c_n), \quad \text{and}\quad B_n=B^X_{r_n}(c_n),
$$
(where the superscript indicates the normed space under consideration for the purpose of defining a ball),
it is clear that $\tilde B_n\cap X=B_n$, and we claim that
$$
\tilde B_n\subseteq \tilde B_{n+1},
\tag 3
$$
for every $n$. In fact, given any $y$ in $\tilde B_n$, we have that
$$
\|y-c_{n+1}\| \leq
\|y-c_n\| + \|c_n-c_{n+1}\| < r_n + {1\over 2^n} = r_{n+1},
$$
by (1) and (2), thus proving the claim. It is not hard to prove that
$$
\bigcup_{n=1}^\infty \tilde B_n = B^{\tilde X}_{r}(a),
$$
so,
if both sets above are intercepted with $X$, we deduce that
$$
\bigcup_{n=1}^\infty B_n = B^{\tilde X}_{r}(a)\cap X.
$$
Observing that $B_n\subseteq B_{n+1}$ by (3), our assumption (ii) implies that $B^{\tilde X}_{r}(a)\cap X$ is an open ball in
$X$, but beware that $a$ is not its center because $a$ is not even in $X$!
Let us therefore write
$$
B^{\tilde X}_{r}(a)\cap X = B^{X}_{r}(b),
$$
for some $b$ in $X$, where we have retained the radius $r$ because the radius of a ball is half its diameter, and it
is clear that the diameter of $B^{\tilde X}_{r}(a)\cap X$ is $2r$.
It follows that $B^{X}_{r_n}(c_n) = B_n \subseteq B^{X}_{r}(b)$ so, by @bof's Lemma (see below), we have
$$
\|c_n-b\|\leq r-r_n \to 0,
$$
as $n\to\infty$, so we see that
$$
b=\lim_{n\to\infty } c_n = a,
$$
contradicting the assumption that $a$ is not in $X$. This concludes the proof.
Lemma (@bof) Let $X$ be a normed space, and pick two elements $c$ and $d$ in $X$, as well as two positive real numbers $r$ and $s$. Assuming that
$B_r(c)\subseteq B_s(d)$,
one has that
$r+\|c-d\|\leq s$.
Proof. This is obvious in case $c=d$. Otherwise note that, for every $t$ in the half-open interval $[0,r)$, one has that
$$
x:= c+{t\over\|c-d\|}(c-d)\in B_r(c).
$$
By assumption $x\in B_s(d)$, so
$$
s>\|x-d\| =
\left\|c-d+{t\over\|c-d\|}(c-d)\right\| =
\|c-d\| + t.
$$
Therefore
$$
s\geq \lim_{t\to r_-} \|c-d\| + t = \|c-d\| + r,
$$
concluding the proof.
