= Solution
{c}
\b[The only <isolated point> of the <subset> $[0,1]$ is $1$.] Indeed,
$$
[1,2)\cap[0,1]=\{1\},
$$
so $\{1\}$ is <open> in the <subspace topology>. If $0\leq x<1$, every basic <neighbourhood> $[a,b)$ containing $x$ also contains a point strictly between $x$ and $\min(b,1)$, hence another point of $[0,1]$. Such an $x$ is not an <isolated point> of the <subset>. Here isolation refers to $[0,1]$ with its <subspace topology>; $1$ is not an <isolated point> of the ambient <Sorgenfrey line>.
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