= Circle and interval are not homeomorphic
The <circle> $S^1$ and the <closed interval> $[0,1]$ are both <compact> and <connected>, so those two properties alone do not distinguish them. Instead examine <connectedness> after removing a point. Every punctured <circle> is <connected>: after rotating the missing point to $(1,0)$, the <continuous map> $t\mapsto(\cos t,\sin t)$ from the <connected> <open interval> $(0,2\pi)$ maps onto it. <Continuous maps> preserve <connectedness>. https://ocw.mit.edu/courses/18-101-analysis-ii-fall-2005/babd982be745679b6d691f78b1c18f53_lectures.pdf[MIT Analysis II, Theorem 1.29]{ref}
If a <homeomorphism> $h:S^1\to[0,1]$ existed, choose $p=h^{-1}(1/2)$. Its restriction would give a <homeomorphism> between $S^1\setminus\{p\}$ and $[0,1]\setminus\{1/2\}$. But the latter splits into two nonempty disjoint <open sets> in its <subspace topology>, $[0,1/2)$ and $(1/2,1]$, contradicting <connectedness> of the punctured <circle>. Removing an endpoint of $[0,1]$ leaves a <connected space>; the proof deliberately chooses the preimage of an interior point. This illustrates how a simple <connectedness> test can rule out every possible <homeomorphism>.
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