Can an open set in $\mathbb{R}^n$ homeomorphic to a non-open set? Suppose $A$ is an open set of $\mathbb{R}^n$ under the ordinary topology and is homeomorphic to $B\subset \mathbb{R}^n$. Is it necessary that $B$ be open too?
 A: Yes, but this is very specific to $\Bbb R^n$ and doesn't hold in arbitrary spaces. It's a highly non-trivial fact (due to Brouwer) called the invariance of domain (see here e.g.); here domain is an old school word for open subset of $\Bbb R^n$.
You cannot really prove it easily without some machinery (Brouwer's fixed point theorem, or some dimension theory; nowadays most will approach it via algebraic topology methods).
A: For $\mathbb R^n$ this is true, as pointed out in comments and other questions, by the (non-trivial) Theorem of invariance of domain:
If $U\subseteq \mathbb R^n$ is open and $f:U\to \mathbb R^n$ is continuous and injective, then $f(U)$ is open. (See here)
So if $f$ is an homeo between an open sub-set and a second sub-set of $\mathbb R^n$, also the second must be open.
When you don't have this theorem, the claim can be false.
Example: Consider $\mathbb R$ with the co-finite topology: $A$ is open if and only if its complement is finite, empty or the whole $\mathbb R$. Now consider $X=\{x\in\mathbb R: x\neq 1\}$ and $Y=\mathbb R\setminus\mathbb Z$, both with the topology induced  as sub-spaces of $(\mathbb R,\text{co-finite})$. Both $X$ and $Y$ are uncountable sets with the cofinite topology, hence homeomorphic.  But $X$ is open and $Y$ not.
