Is the union of an increasing family of balls a ball?

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.

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 $\operatorname{diam}(S)=d$.