Prove that there is no continuous map from $[0,1]$ onto open interval $(0,1)$. I know the property that continuous image of compact space is compact. Here $[0,1]$ is compact and $(0,1)$ is not. Is this logic correct( / enough) or it requires something more than this? And is there any other method to prove this? Please help. Thanks. :)

That is correct. Note that a continuous function $f$ on a compact interval $[a,b]$ attains a maximum $M$ and minumum $m$ and, by the intermediate value theorem,  $f([a,b])=[m,M]$ which is a closed interval.
