 Solution

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 ${\mathbb{R}}_{\mathcal{T}}$.