Qiaochi's answer deals with closed normal subgroups. However, a priori, there could have been non-closed proper normal subgroups in $SU(n)$ (not contained in its center). One can show that this cannot happen "with bare hands" (this is done for instance in Berger's "Geometry" book in the case of $SO(3)$ merely using elementary geometry), or use the following general theorem:
Theorem. Suppose that $G$ is a compact Hausdorff topological group, which is topologically simple, i.e. contains no proper closed normal subgroups. Then $G$ is simple as an abstract group.
See Theorem 9.90 of
Karl H. Hofmann, Sidney A. Morris, The Structure of Compact Groups. A Primer for the Student – A Handbook for the Expert, de Gruyter Studies in Mathematics, Volume 25, 1998.
Addendum. There exist Hausdorff topologically simple groups which are not simple as abstract groups, see section 3 in
George A. Willis, Compact open subgroups in simple totally disconnected groups, Journal of Algebra, Volume 312, Issue 1 (2007) pages 405-417.