A reasoning on compactness

It is true that both intervals $[0,\frac{2}{3})$ and $(\frac12,1]$ are open in $[0,1]$, however this is not the definition of compactness.

Indeed, one has to consider an arbitrary covering of the desired set, and then deduce that there always exists a finite subcovering of our set. Which is not what we have done here since we already chose a specific covering of our set $[0,1]$.