Let $M$ be an $n$-manifold. I would like to show that any open subset $A$ of $M$ is an $n$ manifold. Let $x\in A$. since $M$ is a manifold, there exists a neighborhood $U_x\subseteq M$ of $x$ such that $U_x$ is homeomorphic to an open subset $V$ of $\mathbb R^n$, let $h:U_x\to V$ be such homeomorphism. Consider the restriction of $h$ to $A\cap U_x$. The set $A\cap U_x$ contains the open set $A$ that contains $x$ so $A\cap U_x$ is a neighborhood of $x$. The proof is finished if i could say that $h(A\cap U_x)$ is an open subset of $\mathbb R^n$. this is obvious if $U_x$ were open because in this case i would say that $A\cap U_x$ is open in $M$ and since $h$ is homeomorphism then it is an open map hence $h(A\cap U_x)$ is an open subset of $\mathbb R^n$. But the Problem is that $U_x$ need not be open..
Your reasoning is correct.
Notice that you can assume without loss of generality that $U_x$ is open. After all, the definition of neighborhood is that $U_x$ contains an open subset, say $U_x'$, which contains the point $x$. Then $U_x'$ is an open neighborhood of $x$ which is homeomorphic to an open subset of $\mathbb R^n$, namely $h(U_x')$.