I'll assume that you already have Clauses (B)(2)-(3). We need to thin down once more to get (1). WLOG, we can assume each $\lambda_{\alpha}$ is a successor cardinal strictly above $\sup_{\beta < \alpha} \lambda_{\beta}$. Put $\mu = cf(\lambda) > \aleph_0$. First apply (A) to the family $\{A_{\alpha}: \alpha < \mu\}$ to get $A$ and $X \in [\mu]^{\mu}$ such that for every $\alpha \neq \beta$ in $X$, $A_{\alpha} \cap A_{\beta} = A$. WLOG, $X = \mu$.

Fix $\alpha < \mu$. Put $W_{\alpha} = \bigcup \{A_{\beta, \gamma}: \beta < \alpha, \gamma < \lambda_{\beta}\}$. Then $|W_{\alpha}| < \lambda_{\alpha}$. Observe that for each $\alpha < \mu$, $\{A_{\alpha, \beta} \setminus A_{\alpha}: \beta < \lambda_{\alpha}\}$ is a family of pairwise disjoint sets. So fewer than $\lambda_{\alpha}$ of these intersect $W_{\alpha}$. Throw these away and check that (1) holds.

I'll assume that you already have Clauses (B)(2)-(3). We need to thin down once more to get (1). WLOG, we can assume each $\lambda_{\alpha}$ is a successor cardinal strictly above $\max(\mu^+, \sup_{\beta < \alpha} \lambda_{\beta})$. Put $\mu = cf(\lambda) > \aleph_0$. First apply (A) to the family $\{A_{\alpha}: \alpha < \mu\}$ to get $A$ and $X \in [\mu]^{\mu}$ such that for every $\alpha \neq \beta$ in $X$, $A_{\alpha} \cap A_{\beta} = A$. WLOG, $X = \mu$.

Fix $\alpha < \mu$. Put $W_{\alpha} = \bigcup \{A_{\beta}: \beta < \mu\} \cup \bigcup \{A_{\beta, \gamma}: \beta < \alpha, \gamma < \lambda_{\beta}\}$. Then $|W_{\alpha}| < \lambda_{\alpha}$. Observe that for each $\alpha < \mu$, $\{A_{\alpha, \beta} \setminus A_{\alpha}: \beta < \lambda_{\alpha}\}$ is a family of pairwise disjoint sets. So fewer than $\lambda_{\alpha}$ of these intersect $W_{\alpha}$. Throw these away and check that (1) holds.

Edit: I have added the union of the $A_{\alpha}$'s to $W_{\alpha}$. (1) should hold now.