$\newcommand{\d}{\operatorname{d}}\newcommand{\b}{\mathcal{B}}\newcommand{\t}{\mathcal{T}}\newcommand{\A}{\operatorname{Arcsin}}$I'll let $K$ denote both the set $[0,1)$ and the metric space $(K,\d)$, for $\d:K^2\to K,\,(x,y)\mapsto|\exp(2\pi i\cdot x)-\exp(2\pi i\cdot y)|$. I let also $\t_K$ denote the induced topology on $K$, and $\t_E$ denote the Euclidean topology on $\Bbb R$; analogously I denote by $\b_{K},\b_{E}$ the metric balls w.r.t each space.
I have never done a proof of compactness before, so I am worried that what follows is unrigorous.
It is easily seen that $\d$ is a metric. It can also be seen that, due to the restriction to the interval on $[0,1)$, we have $\d(x,y)=2\sin\pi|x-y|$.
If $\varnothing\neq V\in\t_K$, then for some $x\in V$, $\exists r\gt0$: $\b_K(x,r)\subseteq V$. This ball takes an unusual shape however: $$\b_K(x,r)=\{y\in K:\d(x,y)\lt r\}=\{y\in K:2\sin\pi|x-y|\lt r\}$$If $2\sin\pi|x-y|\lt r$, then $\pi|x-y|\lt\A\frac{r}{2}$ or $\pi|x-y|\gt\pi-\A\frac{r}{2}$, in light of the fact that $|x-y|\in[0,1)$. It follows that, if $\eta(r)=\frac{1}{\pi}\A\frac{r}{2}$, we have: $$\b_K(x,r)=\{y\in K:|x-y|\lt\eta(r)\}\cup\{y\in K:|x-y|\gt1-\eta(r)\}$$
Put: $$\begin{align}\lambda_1&=\min\{y-x,x-y+\eta(r)\}\\\lambda_2&=\min\{x-y,y\}\\\lambda_3&=\min\{1+x-y,y-x+\eta(r)-1\}\\\lambda_4&=\min\{1+y-x,x-y+\eta(r)-1\}\end{align}$$
Suppose $y\in\b_K(x,r)$. If $y$ lies in the first interval, and $y\gt x$, then $y$ is enclosed in $\b_E(y,\lambda_1)$, else if $y\lt x$ it is enclosed in $\b_E(y,\lambda_2)$. If $y$ lies in the second interval, if $y\gt x$ it is enclosed in $\b_E(y,\lambda_3)$, and if it $y\lt x$ than $x$ is is enclosed in $\b_E(y,\lambda_4)$. By "enclosed", I also mean that each of these Euclidean balls is contained in $K$. This holds for $y\neq 0$.
Then if $V\in\t_K,\,x\in V$ we have $\b_K(x,r)\subseteq V$ by definition, and by the above we have $\b_E(y,\lambda)\subseteq V$ for every $0\neq y\in V$ and some $\lambda$ chosen as above. Then $V\in\t_k,\,0\notin V\implies V\in\t_E$.
Now suppose there is an open family $\{V_i:i\in I\}\subseteq\t_K$, where $I$ is an index set, such that $K=\bigcup_{i\in I}V_i$. As $0\in K$, $\exists j\in I:0\in V_j$. Then there exists also some $r\gt0$ such that $\b_K(0,r)\subseteq V_j$. If $V_j=K$, then we are trivially done as the finite subcover may simply be $V_j$. Else, $K\setminus V_j\neq\varnothing$, and $K\setminus\b_K(0,r)=F$ is covered by $\bigcup_{i\in I}V_i$. Now, $K\setminus\b_K(0,r)=\{y\in K:\eta(r)\le y\le 1-\eta(r)\}=[\eta(r),1-\eta(r)]$. By the Heine-Borel theorem, as $\eta(r)$ is bounded $\le\frac{1}{2}$, $F$ is compact in $\t_E$. However, as $\bigcup_{i\in I}V_i$ is an open cover of $F$, and $0\notin F$, we have by the discussion above that that $\{V_i\}$ is analogous to an open cover in $\Bbb R$ and by compactness it admits a finite subcover, call it $G$. Then, taking $V_j\cup G$, we are done.
I feel like this isn't quite right but I don't know how to fix it.