I don't think there is an easy necessary condition, besides the trivial "normal" itself, or some reformulation thereof. What is known is thet $F_\sigma$ (countable unions of closed sets) subsets of normal spaces are normal in the subspace topology, and there are examples of open subspaces of compact Hausdorff spaces that are not normal; even removing a point from such a space can leave a non-normal result. If a normal space $X$ has the property that all open subspaces are normal, then all of its subspaces are normal, so openness is the "test case" , as it were.