Past exam of the topology course of the University of Oxford / 2021 / 1 / e

$X$ is not compact. In the subspace topology on $X=[0,1]$, consider the open coverThe intervals are basic open sets in the ambient Sorgenfrey line, and $\{1\}=X\cap [1,2)$ is open in $X$. Every $x<1$ belongs to one of those intervals, while the singleton covers $1$.

A finite subcover would have to include $\{1\}$. Among its finitely many other members, let $N$ be the largest index. Their union is $[0,1-1/N)$, leaving $1-1/N$ uncovered. Selecting no such member also fails to cover $0$. Hence no finite subcover exists.

$X$ does not have sequential compactness. Take the sequenceEvery subsequence has usual limit of a sequence $1$. If a subsequence converged in the subspace topology of the Sorgenfrey line, it would also converge in the usual subspace topology, since part (c) makes the identity function from the former to the latter continuous. Uniqueness of a limit of a sequence in the usual Hausdorff space therefore forces that limit of a sequence to be $1$.

But $\{1\}$ is an open neighbourhood of $1$ in $X$, and no term $x_n$ belongs to it. Thus no subsequence converges to $1$ in $X$, proving the assertion.