= Solution
{c}
\b[$X$ does not have <sequential compactness>.] Take the <sequence>
$$
x_n=1-\frac1n\qquad(n\geq1).
$$
Every <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.
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