There are no compact minimal surfaces

Or can we say the following?

Since a compact surface must have an elliptic point, if there was a compact minimal surface, then the existence of an elliptic point $p$ would mean $\kappa_1(p)\kappa_2(p)>0$, where $\kappa_1(p)=-\kappa_2(p)$, which would imply $\kappa_1(p)\kappa_2(p)=-\kappa_1(p)^2<0$, contradiction.

Well actually this "proof" uses the crucial fact that compact surfaces have elliptic points, whose proof is similar to the ones given in above. Since Lazywei is asking this question as an exercise in do Carmo's book, and the crucial fact is also an exercise in do Carmo's book,

If a surface $S$ is compact, then every linear functional, such as $f(x)=x_1$, will attain its maximum $M$ somewhere on it. In some neighborhood of the maximum point, $S$ is the image of a (conformal) harmonic embedding $h:U\to \mathbb R^3$, where $U$ is a domain in $\mathbb R^2$. It follows that $f(h)$ attains interior maximum, and is therefore constant. It follows that the intersection of $S$ with the plane $x_1=M$ is both open and closed in $S$. Hence, $S$ is contained in a plane, which quickly leads to a contradiction.