A subset $K \subset \mathbb{C}^n$ is holomorphically convex in some open set $X \subset \mathbb{C}^n$ if the set $\{x \in X : |f(x)| \leq \mathrm{sup}_{y\in K}|f(y)| \quad \textrm{for all holomorphic } f: X \to \mathbb{C}\}$ equals $K$.
Let $W \subset \mathbb{C}^n$ be an open ball centred at the origin and let $B \subset W$ be a closed ball centred at the origin. Is it obvious that $B$ is holomorphically convex in $W$?
Let $B = \{ s\in \Bbb{C}^n,\sum_j |s_j|^2\le r\}$. If $s\not \in B$ then $$f(z)=\frac{1}{\sum_j \overline{s_j}z_j - \sum_j |s_j|^2}$$ is holomorphic and bounded on $B$ and it has a pole at $s$ thus $s\not \in HolomorphicClosure(B)$.
If you want the function to be holomorphic on the larger ball $W$ then look instead at $$F(z)=\sum_{k=0}^K (\frac{\sum_j \overline{s_j}z_j}{ \sum_j |s_j|^2})^k$$ (I'm claiming both criteria, in term of holomorphic and meromorphic functions, are equivalent on such simply connected domains in $\Bbb{C}^n$). When $W$ is a compact complex manifold there are no non-constant holomorphic functions so the meromorphic criterion is more natural to me.