= Solution
Use $\sigma_A(x)$ and $\rho_A(x)$ for <Banach algebra spectrum> and <algebra resolvent set>. An inverse in the smaller <unital algebra> is also an inverse in the larger one, so $\rho_B(x)\subseteq\rho_A(x)$.
If $\lambda\in\rho_B(x)$ and $r=(\lambda1-x)^{-1}\in B$, then for $|\mu-\lambda|\|r\|<1$ the <Neumann series> gives
$$
(\mu1-x)^{-1}=(1+(\mu-\lambda)r)^{-1}r\in B.
$$
Thus $\rho_B(x)$ is <open>. To prove relative closedness, suppose $\lambda_n\in\rho_B(x)$ and $\lambda_n\to\lambda\in\rho_A(x)$. <Continuity> of inversion in $A$ gives $(\lambda_n1-x)^{-1}\to(\lambda1-x)^{-1}$. Since $B$ is <norm> <closed>, the limiting inverse belongs to $B$, so $\lambda\in\rho_B(x)$. Consequently \b[$\rho_B(x)$ is relatively clopen in $\rho_A(x)$].
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