$A$ open and $A \simeq B$ $\Rightarrow$ $B$ open, or $A$ closed and $A \simeq B$ $\Rightarrow$ $B$ closed For 2 topological spaces $(X,T_X)$ and $(Y,T_Y)$, I write $X \simeq Y$ if $X$ and $Y$ are homeomorphic. If $A \subset X$, I always endow $A$ with the subspace topology.
I was led to consider the following assumptions.

(1) $(X,T_X)$ is a topological space such that for all $A$,$B$ $\subset X$
  : $A$ closed and $A \simeq B$ $\Rightarrow$ $B$ closed

It is true if $X$ is discrete or compact Hausdorff, but it's not true if $X = \mathbb{R}^n$, $n \geq 1$ (because $\mathbb{R}^n \simeq (0,1)^n$), or if $X$ has a closed singleton and an other non-closed singleton (like $X=\{a,b\}$ with $T_X = \{ \varnothing, \{a\}, \{a,b\}$).

(2) $(X,T_X)$ is a topological space such that for all $A$,$B$ $\subset X$
  : $A$ open and $A \simeq B$ $\Rightarrow$ $B$ open 

Again, it is true if $X$ is discrete, or if $X = \mathbb{R}^n$ with $n \geq 0$ (invariance of domain), but it's not true if $X$ has an open singleton and an other non-open singleton (like $X=\{a,b\}$ with $T_X = \{ \varnothing, \{a\}, \{a,b\}$).
My question is :

What are the conditions on the space $X$ that guarantee that one of $(1)$ and $(2)$ is true, or at least some examples apart from the ones
  mentionned ?

A topic already exists (Homeomorphic closed subspaces.) that deals whith $(1)$ but they don't give the general answer there. However, they say in this topic that if $X$ is countably compact and first countable, $(1)$ holds : why ?

I doubt that there is a nice answer to either question. However, $(1)$ holds if $X$ is Hausdorff, sequential, and countably compact; every first countable space is sequential, so this generalizes the result mentioned in the comment under the earlier question. (That comment did not specify Hausdorffness, but it is required.)
Suppose that $X$ is Hausdorff, sequential, and countably compact, $A$ is a closed subset of $X$, and $h:A\to B\subseteq X$ is a homeomorphism. If $B$ is not closed in $X$, then $B$ is not sequentially closed, so there is a sequence $\langle x_n:n\in\Bbb N\rangle$ converging to some $x\in X\setminus B$. Let $D=\{x_n:n\in\Bbb N\}$; since $X$ is Hausdorff, $\operatorname{cl}D=D\cup\{x\}$, and $D$ is therefore infinite, closed, and discrete in the subspace $B$. But then $h^{-1}[D]$ is infinite, closed, and discrete in the subspace $A$, which is impossible: $A$, being a closed subset of $X$, is countably compact and therefore contains no such subset.
