Example of two open balls such that the one with the smaller radius contains the one with the larger radius.

Have you considered the trivial $1$ point metric space? Every ball is the same there.

In general, pick any space with an isolated point. Then you can have equal balls around it with different radii.

Also note that every metric space can be bounded by cutting off the distance, i.e. $d'(x,y)=\min(d(x,y),1)$. Then $B(x,r)=B(x,r')$ for any $r,r'\ge1$.

If you are looking for a ball of larger radius strictly contained in a ball of smaller radius, consider the metric space as the interval $[-1,1]$ which is an open ball of radius $\frac43$ around $0$ which strictly contains an open ball of radius $\frac53$ around $1$.

Look at the plane ($\mathbb{R}^2$) in the so-called post-office metric. Denoting the usual norm on the plane by $\|x\|$, we define this metric for two distinct points $x$ and $y$ in the plane as $d(x,y) = \|x\| + \|y\|$, and $d(x,x) = 0$, of course. One easily checks that this is a a metric.

The intuition is that to go from point $x$ to $y$ they need to travel via the "post-office", which is the origin $(0,0)$. One sees that every point $x \neq (0,0)$ is an isolated point, and that open balls around $(0,0)$ are the same as the usual ones.

Now look at $B((0,0), \frac{3}{2})$ and $B((1,0), 2)$. The latter consists of the usual open ball around $(0,0)$ of radius $1$ plus $(1,0)$ itself, which is a proper subset of the former, which has a smaller radius.

If you just want containment, the trivial metric space with a single point is the best example.

If you want that the ball with the larger radius is a proper subset of the one with the smaller radius they will have to have different centers, but then it is rather straight-forward.

Depending on your preferences for examples, either take:

3 points a, b, c with d(a,b) = d(b,c) = 1 and d(a,c) = 2. Then B(b,1.5) = {a,b,c} and B(a,1.6) = {a,b}.

Or take the unit interval [0;1], and then consider the balls B(0.5,0.6) and B(0,0.7).