Let $M$ be a smooth manifold (with dimension $m$, if relevant). Let $B$ be an immersed submanifold of $M$. Let $A$ be an open subset of $B$. Is $A$ an immersed submanifold of $M$? I think so. Please verify:
By definition of $B$ as an immersed submanifold of $M$, there exists a smooth manifold $N$ and an immersion $F: N \to M$ such that $F(N)=B$.
Hopefully, $N$ has a dimension $n$ too, if relevant. (I understand that for manifolds with dimension, immersions imply $\dim N \le \dim M$. I don't know of any definition of immersion where $N$ or $M$ may not have dimension.)
$A$ is an immersed submanifold of $M$ if there exists a smooth manifold $P$ and an immersion $G: P \to M$ such that $G(P)=A$
Choose $P=F^{-1}(A)$ and $G=F|_{F^{-1}(A)}$. This works because:
4.1. $P=F^{-1}(A)$ is a manifold: $F$ is smooth implies $F$ is continuous implies $\tilde F: N \to F(N)=B$ is continuous implies that $P=F^{-1}(A)$ is open in $N$ and thus is a regular/embedded submanifold (of codimension zero) and thus a manifold.
4.2. The image of $G$ is $A$: $G(F^{-1}(A))=F(F^{-1}(A))=A$ since $A \subseteq B = F(N))$.
4.3. $G$ is smooth: This follows by (4.1) because restrictions of smooth maps to (regular/embedded) submanifolds with dimension (The only definition of submanifold I know is where both the submanifold and original manifold have dimensions), including and especially the ones of codimension zero, are smooth.
4.4. $G$ is an immersion: Let $p \in P$. Let $\iota: P \to N$ be the inclusion map, which is proven smooth as part of proving (4.3). Then, by chain rule, $G_{*,p}=(F \circ \iota)_{*,p}=F_{*,p} \circ \iota_{*,p}$. Since inclusions from (regular/embedded) submanifolds are immersions, $G$ is an immersion since compositions of injections are injective.