Show that $S^1$ is not homeomorphic to $[0,1)$ I want to show that $S^1$ is not homeomorphic to $[0,1)$.
When $X$ is homeomorphic to $Y$ then $X-\{a\}$, where $a \in X$ is homeomorphic to $Y-\{f(a)\}$. Assuming $S^1$ and $[0,1)$ are homeomorphic, I remove a point $a$ other than $0$. We see that $[0,1)-\{a\}$ is homeomorphic to $S^1-\{f(a)\}$. But  $S^1-\{f(a)\}$ is connected while $[0,1)-\{a\}$ is not. This is a contradiction which proves the claim.
However when I remove $0$, We see that  $ [0,1)-\{0\}$ is homeomorphic to $S^1-\{f(0)\}$ which is in fact true. This shows $S^1$ is indeed homeomorphic to $[0,1)$ opposite of what I am trying to show. So I am not sure what has gone wrong in these reasonings.

Now I understand your confusion, actually it is more like a matter of logic:
You were assuming the statement $P$ (that they are homeomorphic) at the very beginning, your aim is to deduce that $\neg P$ (that they are not homeomorphic) but you obtain by another way that the  second argument that they are homeomorphic (that $[0,1)\setminus\{0\}$...), this does NOT show that it is in fact $P$ because you have had assumed at the beginning, this is not a way of showing a statement by assuming at the very beginning.
