In a book I'm reading, it's written that in every normed spaces, balls are convex. In every metric space, ball are bounded.
Since they didn't mention that in metric spaces balls are convex, could it happen that a ball is not convex in a metric space ? In the french page of wikipedia, they also claim the same statement. But I didn't see in the english page...
Of course, the comment by Lord Shark is correct, but here is an example that shows that balls need not be convex.
Consider the space $L^p( \{ \{0,1\} )$, for $0 < p < 1$. The $L^p$ spaces in general are the space of all functions on a set that are $p$-times Lebesgue integrable, i.e. when you integrate them to the $p$th power, you get a finite number.
For us, this is a sort of degenerate, trivial example, as to 'integrate' a function on a set with two points is just to add up the values. So being $p$ times integrable therefore means that if $f$ is a function on these two values, it is simply a pair of numbers, and $\int f^p$ is simply $|x|^p + |y|^p$.
Now this space is not normed, but it does have a metric. This is given by $\int |f-g|^p$, which is of course just $|x_1 - x_2|^p + |y_1 - y_2|^p$. Let's consider the unit ball at the origin, so we set $x_2 = y_2 = 0$. This ball is the set of all functions, i.e. pairs of points for which $|x|^p + |y|^p \leq 1$.
We can graph this in Wolfram Alpha, say with $p =1/2$
Here's a graphic.
And we see that this region is not convex! The segment connecting the points leaves the set.
When you know more measure theory and analysis, you will be able to make the corresponding generalization to $L^p(X)$, for $X$ a more general type of set, and the same kind of observation will be true.