 Solution

$X$ 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.