Are manifold subsets submanifolds?
Solution 1:
No, this is very very false. For instance, let $B$ be $\mathbb{R}$ with its usual smooth manifold structure, and let $A$ be $\mathbb{R}$ with a smooth manifold structure given by picking a bijection $\mathbb{R}\to\mathbb{R}^2$ and pulling back the usual smooth manifold structure on $\mathbb{R}^2$. Then $A$ is certainly not an embedded submanifold of $B$, since it has larger dimension. Indeed, the inclusion map $A\to B$ cannot even be continuous.
Even if you assume $A$ has the subspace topology, it is still very false. For instance, in the example above, you can instead pick a homeomorphism $\mathbb{R}\to\mathbb{R}$ that is not a diffeomorphism and pull back the usual smooth manifold structure of $\mathbb{R}$ to a new one and call it $A$. Then the inclusion map $A\to B$ will be a homeomorphism but not a diffeomorphism.
The key thing to understand here is that being a manifold is not a property of a set. It's an additional structure you can put on a set. All that that $A\subseteq B$ tells you is that every element of $A$ happens to be an element of $B$; it tells you nothing at all about their manifold structures, which could be totally unrelated. (The same thing happens with rings: if $A$ and $B$ are rings with $A\subseteq B$, then there is no reason at all to think that $A$ is a subring of $B$, because the ring operations of $A$ are probably totally different from those of $B$.) Being a smooth manifold is similarly not a property of a topological space, but an extra structure you can put on it.
As for your proposed proof, all three of your claims are wrong as shown by the example above. You gave no justification for claim 1 or claim 2 ("inclusions are the prototype of immersions" is just a vague slogan that has no meaning in a proof). For claim 3, to prove $\iota$ is an embedding you need to prove it is a homeomorphism from $A$ to $\iota(A)$ with the subspace topology from $B$, and you have no reason to believe that topology is the same as the given topology on $A$.