Let $B=\{z\in \mathbb{C}^N: |z|<1\}$ be the (Euclidean) open unit ball in $\mathbb{C}^N$, let $e_1=(1,0,\ldots,0)$, and let $\Omega=\{z\in \mathbb{C}^N: |z+e_1|<2\}$ be the open ball of radius 2 centered at $-e_1$, so that $B\subset \Omega$. For a domain $G\subset \mathbb{C}^N$, let $H^\infty(G)=\{f:\mathbb{C}^N\to \mathbb{C}: f$ is analytic on $G$ and $\sup_{z\in G}|f(z)|<\infty\}$. Then $H^\infty(\Omega) \subset H^\infty(B)$, and in the topology of uniform convergence on compact sets, $H^\infty(\Omega)$ is dense in $H^\infty(B)$ since $H^\infty(\Omega)$ contains the polynomials.
In the topology of uniform convergence on compact sets, is the ball (of $H^\infty(\Omega)$) denoted by $X_\Omega=\{f\in H^\infty(\Omega): \sup_{z\in\Omega}|f(z)|\leq 1\}$ dense in the ball $X_B=\{f\in H^\infty(B): \sup_{z\in B}|f(z)|\leq 1\}$ of $H^\infty(B)$?
No. WLOG $N=1$. Let $D$ be the disk with center $1/2$ and radius $3/2$, so $$B\subset D\subset\Omega.$$Writing $$||f||_E=\sup_{z\in E}|f(z)|,$$"Cauchy's Inequality" for the first derivative says
If $f_n\in H^\infty(\Omega)$ and $f_n(z)\to z$ uniformly on compact subsets of $B$ then $f_n'(1/2)\to1$; hence we must have $\liminf||f_n||_D\ge3/2$.