When $A$ is a subset of a topological space $X$ then the closure of $A$ is the smallest closed set of $X$ containing $A$. Analogously, the interior of $A$ is the largest open set of $X$ contained in $A$.
In cases where $A$ is a subset of more than one topological space of interest, as in the example $A\subseteq Y\subseteq X$ where $Y$ is a subspace of $X$, it is customary to note with respect to which topology the closure/interior is taken. i.e $Cl_X(A), Int_Y(A)$ etc.
This specification is important as even when $Y$ is a subspace of $X$ the closure of $A$ with respect to $Y$ may be different from the closure with respect to $X$. Same for interior. This happens because while the topologies of $Y$ and $X$ are intimately related, they are not the same topology, and hence have different closed and open subsets.
In your specific example, for instance, while $[1,9]$ is obviously not an open subset of $X$, it is an open subset of $Y$ (because it is the entire space). Therefore $Int_X(Y)$ is indeed $(1,9)$, but $Int_Y(Y)$ is $[1,9]$. Constructing an example where the closures differ is also possible, and not too difficult.
My first thought was that this is a mistake, because we should view them as topological spaces by themselves, not from an outside view, from an ambient space
Well, first of all we can treat them however we want. What exactly can stop imagination, especially in maths, am I right?
Sometimes we do treat them like separate spaces, sometimes we don't. Depends on the context, on what we are trying to achieve.
But in your example I'd say that the ambient space is here only to give definition of topology for $X,Y\subseteq\mathbb{R}^n$.
since these properties change when we consider different topologies.
Different topologies of what? Ambient space? Because $X,Y$ have to have fixed topologies, they cannot change (otherwise we would not be able to prove anything about them). So are you asking about different properties of $X,Y$ with respect to some other ambient space $Z$ they both embed to? Well, such property is not a topological property by definition. The term has a concrete meaning btw.
So for example $X$ being open in $Z$ is not a topological property. But properties of $X$ like: cardinality, separation axioms, compactness, connectedness, and so on (the list is long, see the wiki on topological property) are topological properties. If $X$ is connected, then $X$ is always connected, regardless of any space $Z$ and any embedding of $X$ into $Z$.
Well, this is not actually true, since a set can be neither open or close in $\mathbb{R}^n$, but it is in its own subspace topology.
Being open/closed is not a topological property. In fact there is no "being open/closed" property. There is "being open/closed in [somewhere]" property. This property requires a pair "space, subspace", so of course we cannot think about them separately by definition. It just doesn't make sense.
But you are asking about "being homeomorphic". This is a very different beast and it does not depend on any ambient space.
Bijection is not a topological property.
Well, the term does not apply to topological spaces at all. It is a property of a function. Did you mean that "cardinality" is not a topological property? Of course it is. After all every homeomorphism is a bijection.
Perhaps you wanted to say that cardinality is not a complete invariant. Meaning two spaces may be equinumerous, yet they don't have to be homeomorphic. But that is not really surprising, there are lots of (all?) topological invariants which are not complete. So for example connectedness. Are any two connected subsets of $\mathbb{R}^n$ homeomorphic? Of course not. Yet, this is a topological invariant. Meaning if two spaces are homeomorphic then they have to share it.
My final question is: Is because of the theorem above that we can do what we do (i.e, look up for contradictory topological properties of subspaces viewed from the ambient to conclude there is not homeomorphism between subspaces) ? Or could we do this since the beginning and I am missing something?
The theorem is completely irrelevant to the topic to be honest. Yes, we could do this from the beginning. I think that the biggest mistake you've made is that you focused too much on "being open/closed" which is not even a property of a topological space (but rather a pair of spaces).
Best Answer
$A=S^1\setminus \{p\} \subseteq X= S^1$ (for any $p \in S^1$) is homeomorphic to $B=(0,1) \subseteq Y = [0,1]$. But their respective closures $X$ and $Y$ are not.
More trivially: $A = (0,1) \subseteq X=\mathbb{R}$ and $B = Y = \mathbb{R}$, where $\overline{B} = B$ but $\overline{A}$ becomes compact.