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

Yes, $[0,1]$ is closed. Its complement is $(-\infty ,0)\cup (1,\infty )$. Every $x<0$ has the basic open neighbourhood $[x,0)$ contained in this complement, and every $x>1$ has $[x,x+1)$ contained in it. Hence the complement is open in ${\mathbb{R}}_{\mathcal{T}}$.

No, $[0,1]$ is not open. Every basic neighbourhood $[a,b)$ of $1$ has $b>1$, so it contains points larger than $1$. Therefore $1$ has no open neighbourhood contained in $[0,1]$.

The interval $[0,1)$ is a basic open set contained in $[0,1]$. The remaining point $1$ is not an interior point, by the preceding argument. Thus

The only isolated point of the subset $[0,1]$ is $1$. Indeed,so $\{1\}$ is open in the subspace topology. If $0\le x<1$, 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.