Example of a metric space where diameter of a ball is not equal twice the radius
Consider the discrete metric $d$ on a set $X$:
$$d(x,y)=\begin{cases} 0,&\text{if }x=y\\ 1,&\text{if }x\ne y\;. \end{cases}$$
Consider the ball of radius $r=1/2$ centered at $x$
Then $B(x,r)=\{x\}$
Now by definition, $\operatorname{diam} A = \sup\{ d(a,b) : a, b \in A \}$
Applying it to our case where $A=B(x,r)$, we have diameter of $A$ equal to $0$
How about the metric space $[0,\infty)$ with the standard metric? The diameter of the ball $B(0,1)$ is $1$, not $2$ (which would be twice its radius)
This is an elaboration of @user10354138's answer.
Consider a function $d: X \times X \to [0,\infty)$ which is definite, symmetric, and satisfies the ultrametric inequality $$ d(x,y) \leq \max\{ d(x,z),d(z,y) \}\qquad \text{for all } x,y,z \in X \tag{$*$} $$ instead of the usual triangle inequality. Since $\max\{ d(x,z) , d(z,y) \} \leq d(x,z) + d(z,y)$, we see that that $d$ is indeed a metric.
A space $X$ equipped with a metric satisfying $(*)$ has a topology that behaves counter-intuitively in many ways. For instance, every point in a ball is its centre. Precisely, if $x \in X$ and $r > 0$, then for every $y \in B(x,r)$, we have $B(x,r) = B(y,r)$.
To prove this, suppose $y \in B(x,r)$ as above. Suppose $z \in X$ such that $z \in B(y,r)$. We want to show that $z \in B(x,r)$. By the ultrametric inequality and the definition of an open ball, we have $$d(x,z) \leq \max\{ d(x,y), d(y,z) \} < \max\{ r, r \} = r.$$ So, $B(y,r) \subseteq B(x,r)$. Since $x$ and $y$ were arbitrary, the other containment is also true, so $B(x,r) = B(y,r)$ as was to be shown.
What does this mean in the context of diameters? Well, $diam(B(x,r)) = \sup\{ d(x,y) : x,y \in B(x,r) \}$. Since every point of $B(x,r)$ is its centre, $d(x,y) < r$ for all $x,y \in B(x,r)$. Hence, $diam(B(x,r)) \leq r$, so the diameter is no more than the radius!
In fact, there does not exist any open ball in this metric space whose diameter is twice its radius!
Of course, all this theory would be pointless if it turned out that there is no metric satisfying $(*)$. Thankfully, there is indeed a bunch of metrics that satisfy $(*)$, and I will produce some of them below.
Let $X = \mathbb{Q}$, and let $p \in \mathbb{N}$ be prime. Write $m/n \in \mathbb{Q}$ as $$ \frac mn = p^k \frac{m'}{n'}, $$ where $k \in \mathbb{Z}$ and $\gcd(m',p) = 1 = \gcd(n',p)$. This can be done because $\mathbb{Q}$ is the quotient field of $\mathbb{Z}$, which is a unique factorisation domain.
By an absolute value on a field $\mathbb{F}$ we shall mean a function $| \cdot | : \mathbb{F} \to [0,\infty)$ that is definite, multiplicative, and satisfies the triangle inequality. We define the $p$-adic absolute value $| \cdot |_p$ on $\mathbb{Q}$ by $$ \left| \frac mn \right|_p = p^{-k}, $$ where $m/n = p^k (m'n')$ as above. It is easy to check that this satisfies all the properties of an absolute value. In fact, it satisfies an inequality stronger than the triangle inequality, very reminiscent of $(*)$: $$ | x + y |_p \leq \max\{ |x|_p,|y|_p \} \qquad \text{for all } x,y\in \mathbb{Q}.\tag{$\dagger$} $$ This is also called the ultrametric inequality.
Now, any absolute value induces a metric by $d(x,y) = |x - y|$. The metric induced by the $p$-adic absolute value is called the $p$-adic metric, and thanks to $(\dagger)$ the $p$-adic metric satisfies $(*)$. So, for every prime $p \in \mathbb{N}$, you have a metric on $\mathbb{Q}$ for which the diameter of a ball is not equal to twice the radius.
The $p$-adic metric is actually an important object in number theory, so $(\mathbb{Q},| \cdot |_p)$ is an example of a space which arises naturally and is a prototypical answer to your question.