Given $[0,1]$ a closed interval on $\mathbb{R}$, we know that $[0,1]$ is compact and $\mathbb{R}$ is not, so these two spaces are not homeomorphic to each other.
But homeomorphic perserves cardinality. Since there is no homeomorphism, does that mean the cardinality are diferent between the two sets?
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$f: X \to Y$ homeomorphism $\implies$ $f:X \to Y$ bijective $\implies$ $\vert X \vert=\vert Y\vert$ but the converse does not hold in general. To see that use your counterexample.
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When considering the question of whether two sets are homeomorphic, we want to make sure that our map is a bijection, but also that it is continuous and has a continuous inverse.
These sets actually have the same cardinality, you can verify this by checking that $\mathbb{R}$~$(0,1)$~$[0,1]$ using the schroeder-bernstein theorem.
The issue arises when considering topological invariants like compactness. By the heine-borel theorem, every closed and bounded subset of euclidean space is compact. $[0,1]$, is both closed and bounded, while $\mathbb{R}$ is unbounded, and hence not compact. Since the continuous image of a compact set must be compact, we cannot have a continous bijection from $[0,1]$ to $\mathbb{R}$.
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To check whether two sets $A,B$ have the same cardinality, you have to just check whether there are injections $f:A\to B$ and $g:B\to A$. If there is, then there exists a bijection $h:A\to B$ which ensures that the two sets have the same cardinality, that is Schroder's theorem. There is no topological structure at play when it comes to checking cardinality, the bare set structure is enough. So you do not need homeomorphism.
For the two sets $A=[0,1]$ and $B=\mathbb{R}$, it is easy to find two such injections, namely $f(x)=x$ and $g(x)=1/(1+e^x)$.
The cardinalities of two sets $X$ and $Y$ are the same iff there exists a bijection between them.
The topological spaces $X$ and $Y$ are homeomorphic iff there exists a bijection between them that is continuous in both directions.
As you can see, homeomorphicity is a much stricter requirement, so it shouldn't come as a surprise that there are sets that have the same cardinality without being homeomorphic.