Why is $NP \subseteq P \implies NP^A \subseteq P^A$ false?
Your statement holds for all oracles $A$ iff $\mathsf{P} \neq \mathsf{NP}$. Indeed, if $\mathsf{P} \neq \mathsf{NP}$ then your statement vacuously holds. Conversely, if $\mathsf{P} = \mathsf{NP}$, then your statement fails for a random oracle $O$, since such an oracle satisfies $\mathsf{P}^O \neq \mathsf{NP}^O$ almost surely.

A similar argument shows that if $A$ is a $\mathsf{PSPACE}$-complete language then your statement holds, since in that case $\mathsf{P}^A = \mathsf{NP}^A$ is known to hold.

The proof that $\mathsf{P} \subseteq \mathsf{NP}$ relativizes, which means that it works even in the presence of an oracle. That is, the proof only uses arguments that keep holding even when the Turing machines are allowed access to an oracle. In all other cases, you cannot conclude from a statement of the form $\mathsf{A} \subseteq \mathsf{B}$ that $\mathsf{A}^O \subseteq \mathsf{B}^O$ for all oracles $O$. The two statements are about two different objects: the first one is about Turing machines without oracle access, and the second one is about Turing machines with oracle access to $O$.

This actually happens in practice, as mentioned in the comments: $\mathsf{IP}=\mathsf{PSPACE}$, but there are oracles $O$ relative to which $\mathsf{IP}^O \neq \mathsf{PSPACE}^O$.