Yes, every open ball is connected.

Suppose the open ball $B(a,r)$ is disconnected: $B(a,r) = U \cap V$ where $U$ and $V$ are nonempty, open and disjoint, and $a \in U$. Since $\overline{V}$ is compact, there is a point $v \in \overline{V}$ whose distance $s = d(a,v)$ to $a$ is minimal. Since $V \subset B(a,r)$, $s < r$ and $v \in B(a,r)$. Note that $U \cap \overline{V} = \overline{U} \cap V = \emptyset$, so $v \in V$ and $v \notin \overline{U}$. Thus we have $v \in B'(a,s)$, but $B(a,s) \subseteq U$ so $v \notin \overline{B(a,s)}$, contradicting the assumption $\overline{B(a,s)} = B'(a,s)$.

Yes, every open ball is connected.

Suppose the open ball $B(a,r)$ is disconnected: $B(a,r) = U \cup V$ where $U$ and $V$ are nonempty, open and disjoint, and $a \in U$. Since $\overline{V}$ is compact, there is a point $v \in \overline{V}$ whose distance $s = d(a,v)$ to $a$ is minimal. Since $V \subset B(a,r)$, $s < r$ and $v \in B(a,r)$. Note that $U \cap \overline{V} = \overline{U} \cap V = \emptyset$, so $v \in V$ and $v \notin \overline{U}$. Thus we have $v \in B'(a,s)$, but $B(a,s) \subseteq U$ so $v \notin \overline{B(a,s)}$, contradicting the assumption $\overline{B(a,s)} = B'(a,s)$.