Exercise: Let $X$ be the unit circle in $\mathbb{R}^2$; that is $X=\{(x,y):x^2+y^2=1\}$ and has the subspace topology.
Deduce that $X\ncong (0,1)$ and $X\ncong[0,1]$.
Since the exercise uses the term "deduce" I thought there must be a quick way to show $X(0,1)\ncong (0,1)$and $X\ncong[0,1]$. I figured out that both are connected so that property would be preserved. I was thinking about cardinality. I know that either $X\ncong (0,1)$ or $X\ncong[0,1]$. once we would have $(0,1)\ncong [0,1]$ which is absurd.
Question:
How should I prove that $X(0,1)\ncong (0,1)$ and $X\ncong[0,1]$?
Thanks in advance!
The symbol "$X(0,1)$" doesn't make much sense. So I assume you've simply meant "$X$". Note that typically the $1$-dimensional unit sphere is denoted by $S^1$.
Usually you show that two spaces are not homeomorphic (or homotopic, I'm not sure what "$\ncong$" means) by showing that they don't share some invariant. One such invariant is connectedness but as you've noted yourself all those topological spaces are connected so it isn't good.
Another such invariant is compactness. Since $(0,1)$ is not compact then it cannot be homeomorphic to $S^1$. Unfortunately $[0,1]$ is compact.
So lets dig further. There is a topology entity known as (non) cut-point. More or less if $X$ is connected then a point $x\in X$ is a cut-point if $X\backslash\{x\}$ is not connected. With that we have: if two spaces are homeomorphic then they have the same number of cut-points. They also have the same number of non cut-points (I encourage you to prove both statements yourself). It can be easily seen that $(0,1)$ has zero non cut-points, $[0,1]$ has two non cut-points (namely $0$ and $1$) and every point of $S^1$ is a non cut-point so there are infinitely many non cut-points on $S^1$. Therefore these are pairwise non-homeomorphic spaces.
If "$\ncong$" is supposed to mean "not homotopy equivalent" then things became more difficult. Because (non) cut-points is a topological invariant but not a homotopy invariant. For homotopy case contractiblity is the invariant you would look for. That's because both $(0,1)$ and $[0,1]$ are contractible while $S^1$ is not. Unfortunately showing that $S^1$ is not contractible is not trivial at all. The proof most likely requires some algebraic topology (e.g. homotopy or homology groups).