Hartogs figure not holomorphically convex

Given $$0 < a, b < 1$$, consider the Hartogs figure $$H$$ given by $$\begin{equation*} H = \{ (z,w) \in \mathbb{D}\times \mathbb{D} \ \ | \ \ |z| > a \} \cup \{ (z,w) \in \mathbb{D} \times \mathbb{D} \ \ | \ \ |w| < b \}. \end{equation*}$$ It is well known that $$H$$ is not a domain of holomorphy; any holomorphic function on $$H$$ is actually holomorphic on the whole of $$\mathbb{D}\times\mathbb{D}$$. Thus, by the well established equivalence between domains of holomorphy and holomorphically convex domains $$H$$ is not holomorphically convex. However, is it possible to prove that $$H$$ is not holomorphically convex straight from the definition without using any equivalent statements nor known facts about $$H$$?

Recall the definition of holomorphic convexity: a domain $$U$$ is said to be holomorphically convex if for every compact subset $$K \subset U$$, the holomorphic convex hull $$\hat{K}_U$$ is also compact in $$U$$. Here the holomorphic convex hull is $$\begin{equation*} \hat{K}_U = \{ z \in U \ \ | \ \ |f(z)| \leq \sup_{\zeta \in K}|f(\zeta)| \ \ \forall f \in \mathcal{O}(U) \}. \end{equation*}$$

In the end you will need to know something about the functions in $$\mathcal{O}(H)$$. Specifically we would need their behaviour near some point on the boundary $$\partial H$$, which would tell us that $$H$$ is not a d.o.h. However since we get the extension to $$\mathbb{D}^2$$ directly by the Cauchy integral formula, it is a lot easier using this.
If we know that $$\mathcal{O}(H) = \mathcal{O}(\mathbb{D}^2)$$ the result follows directly. Consider the Reinhardt domain $$K= \{(z,w) \in \mathbb{C}^2 \; | \; |z|=r_1,\; |w|=r_2\}$$ for real numbers $$r_1,r_2$$.
We know that holomorphic functions on a polydisc $$\Delta_{r_1}\times \Delta_{r_2}$$ have maximum modulus on the distinguished boundary $$\Gamma = \partial \Delta_{r_1}\times \partial \Delta_{r_h}$$ by the maximum principle. Thus the holomorphic hull is $$\begin{equation} \hat K = H \cap \{(z,w) \in \mathbb{C}^2 \; | \; |z| \le r_1,\; |w| \le r_2\}. \end{equation}$$ In particular if we pick $$a, $$b we get a holomorphic hull that is not relatively compact in $$H$$.