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This proof feels to easy. Suggestions?

PROOF: We note that a filter $F$ is principal iff $\land$$F$ $\in$ $F$. (We define $\land$$F$ = $\land$$\{$$G$: $G$ $\in$ $F$$\}$) Suppose that $L$ is a finite lattice, and let $F$ be a filter on $L$. We aim to show $\land$$F$ $\in$ $F$. Since $L$ is finite, $L$ is complete, so that for every subset of $L$ meet and join are defined; hence by definition $\land$$F$ exists. Hence $\land$$F$ $\in$ $F$, and F is principal, as claimed. END PROOF.

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    You can't just use completeness, because there are complete lattices where not all filters are principal. – 2017-01-11

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This is almost correct, but you haven't explained why $\bigwedge F\in F$. Just because $\bigwedge F$ exists in $L$ does not automatically mean it must be an element of $F$. You need to again use the fact that $F$ is finite, so $\bigwedge F$ can be constructed by repeatedly taking binary meets using the elements of $F$. Since $F$ is closed under binary meets, this gives that $\bigwedge F\in F$.