Suppose there exists a non-trivial $n$-dimensional unitary representation of $G$ i.e. a homomorphism of Lie groups from $G$ to $SU(n)$. Then, since $\su(n)$ is a simple Lie algebra, the induced map from $\g \rightarrow \su(n) $ is an isomorphism. Hence, the map from $G$ to $SU(n)$ has discrete kernel and is surjective (as $SU(n)$ is connected).This implies that $G$ is a covering group of $SU(n)$.
However, note that $SU(n)$ is a compact group, and the fundamental group of compact groups is finite. Hence, $G$ is a fiber product of compact $SU(n)$ and a finite (and hence compact) group. Hence, $G$ is compact, by a variation of Tychonoff's theorem of products of compact spaces. This is a contradiction. Hence, $G$ does not have a nontrivial finite dimensional unitary representation.
EDIT: I made a pretty big mistake in the above paragraph. The map from $\g \rightarrow \su(n)$ is only injective, not surjective. Hence, we only get a covering map from $G$ onto a connected subgroup of $SU(n)$. This shows that the existence of a unitary representation of a connected, noncompact, "simple" Lie group $G$ is equivalent to the existence of a noncompact, connected, "simple" subgroup of $SU(n)$ for some $n.$ I can't think of an explanation for why such a subgroup does not exist.