Is a ball always connected in a connected metric space?

It is easy to see the answer is no. Take the unit circle in the plane, with the metric inherited from the Euclidian metric in the plane. Then remove the "top" point (coordinates $(0,1)$). Then a "ball" (small disk) around a point close to "the top", restricted to the "circle with the top removed," will be disconnected. Here the metric space is the circle with the top point removed.

No. The Knaster-Kuratowski fan is a connected subspace of the plane that becomes totally disconnected when a certain point is removed, so open balls centred at the other points cannot be connected if they are small enough to exclude the explosion point.

Just take a plane without a segment, and consider a small ball near the center of the segment. There's no need to get complicated stuff.

There are even complete path-connected metric spaces that contain a point $x$ such that no ball around $x$ is connected, for example

$$ \{\langle x,0\rangle \mid x\ge 1\} \cup \{\langle 1,y\rangle \mid 0\le y\le 1 \} \cup \{\langle x,\tfrac1x \rangle \mid x\ge 1 \} \cup \bigcup_{n=3}^\infty \{ \langle x,\tfrac1n \rangle \mid 2 \le x \le n \} $$

No ball of finite radius around $\langle 2,0\rangle$ is connected.

And here's a sketch of a complete and connected (but not path-connected) subspace of $\mathbb R^2$ where no ball whatsoever, around any point, is connected:

Each of the gaps in the Cantor set on the left will eventually be contracted -- but the smaller the gap, the farther to the right will this happen.

On the positive side, if $(X,d)$ is a length metric space then every open ball $B(a, r)\subset (X,d)$ is path connected. (Every closed ball is connected as well, since, in this situation, each closed ball is either a singleton or is the closure of an open ball and the closure of a connected subset in a metric space is connected.)

Edit. The proof is utterly straightforward: Let $B(a,R)$ be an open ball centered at $a$ of radius $R$. For each point $x\in B(a,R)$, for every $\epsilon>0$ there exists a path $p_{ax}$ (connecting $a$ to $x$ of length $< d(x,a)+\epsilon$. Taking $\epsilon:= R- d(a,x)$, we obtain that this path is entirely contained in the open ball $B(a,R)$. Hence, $B(a,R)$ is path connected.