Is the union of an increasing family of balls a ball? Let $M$ be a metric space and let $\mathscr B$ be a family of open balls in $M$ whose radii are bounded. Assuming that $\mathscr B$ is totally ordered by inclusion, is the union of all members of $\mathscr B$ an open ball?
I believe this is false for the rational numbers, but what if $M$ is complete? Does it hold in every normed space?  Every Banach space?  What about $\mathbb R^n$?

My attempt at the case $M=\mathbb R^n$:
Let $R$ be the set of real numbers formed by the radii of the balls in $\mathfrak{B}$ and
let $r=\sup R$.  If $r\in R$ then the answer is obvious,  so let us suppose otherwise.
One may then find an increasing sequence
$\{r_i\}_i$ in $R$ converging to $r$.  The centers $c_i$ of the corresponding balls $B_i$  must form a converging
sequence (this seems awful obvious,  but might require a rather lengthy proof) so let $c$ be the limit.
I guess it should  also be  clear  that $\bigcup \mathfrak B =  \bigcup_i B_i$,  and it seems reasonable to  try to prove that  this set coincides with
the open ball $B_r(c)$,  centered at $c$, with radius $r$.  Regarding the inclusion
$$
  \bigcup_i B_i\subseteq  B_r(c),
  $$
let $x$ belong to the left-hand-side set,  so there exists some $i_0$ such that $x\in B_{i_0}$ and, since the balls are
increasing, this should also hold for all $i>i_0$.   In other words
$$
  \|x-c_{i}\|<r_{i},\quad \forall i\geq i_0.
  $$
Taking the limit as $i\to\infty$,  one concludes that
$$
  \|x-c_{i_0}\|\leq r_{i_0}.
  $$
Among many, this is one of the main outstanding points!  How to get "$<$" instead of "$\leq $"???
 A: To answer your first question, it's not necessarily true in a complete metric space that the union of a chain of open balls is a ball. Here is a counterexample.
Let $M=\{a_i:i\in\mathbb N\}\cup\{b_i:i\in\mathbb N\}\cup\{c\}$ with the following metric:
$d(a_i,a_j)=1$ if $i\ne j$;
$d(b_i,b_j)=2$ if $i\ne j$;
$d(a_i,b_j)=1$ if $j\le i$;
$d(a_i,b_j)=2$ if $j\gt i$;
$d(a_i,c)=d(b_i,c)=2$.
The triangle inequality holds, since all nonzero distances are $1$ or $2$.
The metric is complete, since every Cauchy sequence is eventually constant.
Let $\mathscr B=\{B_n:n\in\mathbb N\}$ where
$B_n=\{x\in M:d(a_n,x)\le1\}=\{x\in M:d(a_n,x)\lt2\}=\{a_i:i\in\mathbb N\}\cup\{b_i:i\le n\}$.
S0 $\mathscr B$ is a chain of open balls and a chain of closed balls. The union $\bigcup\mathscr B=M\setminus\{c\}$ is not a ball because, for each point $x\ne c$, there is a point $y\ne c$ such that $d(x,y)=d(x,c)=2$.
Regarding your other questions. I'm going to guess that it's true for Banach spaces, false for incomplete normed spaces.
A: The answer is affirmative for Banach spaces:
Theorem. If $S$ is a nonempty bounded open set in a Banach space such that, for any positive number $d\lt\operatorname{diam}(S)$, the set $S$ contains a ball of diameter $d$, then $S$ is an open ball.
Proof. Let $d=\operatorname{diam}(S)$ and $r=\frac12d$. Choose a sequence $B_1,B_2,B_3,\dots$ of open balls $B_n\subseteq S$ such that $\operatorname{diam}(B_n)\to d$. Let $d_n=\operatorname{diam}(B_n)$, let $r_n=\frac12d_n$, and let $c_n$ be the center of $B_n$, so that $B_n=B_{r_n}(c_n)$. Note that
$$r_m+\|c_m-c_n\|+r_n\le d,$$ i.e.,
$$\|c_m-c_n\|\le d-r_m-r_n\le\max\{d-d_m,d-d_n\}.$$
Hence $c_1,c_2,c_3,\dots$ is a Cauchy sequence and converges to a point $c$. I claim that $S=B_r(c)$.
Claim 1. $B_r(c)\subseteq S$.
Proof. Suppose $x\in B_r(c)$, so $\|x-c\|=r-\varepsilon\lt r$. Choose $n$ so that $\|c_n-c\|\lt\frac\varepsilon2$ and $r_n\gt r-\frac\varepsilon2$. Then
$$\|x-c_n\|\le\|x-c\|+\|c-c_n\|\lt(r-\varepsilon)+\frac\varepsilon2=r-\frac\varepsilon2\lt r_n,$$
so $x\in B_n\subseteq S$.
Claim 2. $S\subseteq B_r(c)$.
Proof. Assume for a contradiction that $x\in S$ and $\|x-c\|\ge r$. Since $S$ is open, there is a point $y\in S$ with $\|y-c\|=r+\varepsilon\gt r$. Choose a point $z\in B_r(c)$, antipodal to $y$, with $\|z-c\|=r-\frac\varepsilon2$, so that
$$\|y-z\|=\|y-c\|+\|z-c\|=(r+\varepsilon)+(r-\frac\varepsilon2)=d+\frac\varepsilon2\gt d.$$
So $\|y-z\|\gt d$. Since $y\in S$, and $z\in B_r(c)\subseteq S$ by Claim 1, this contradicts the fact that $\operatorname{diam}(S)=d$.
A: 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.
