Is Not-STCON is NL-Complete?
Your proof doesn't work, since you're using an oracle reduction rather than a many-one reduction.

To show that a problem $A$ is NL-complete, you need to show that for every $S \in \mathsf{L}$ there is a logspace-computable function $f_S$ such that $x \in S$ iff $f_S(x) \in A$ (that's the definition); it's enough to show such a reduction for a single NL-complete problem $S$ (exercise). You show instead that $x \in S$ iff $f_S(x) \notin A$. 

Showing that nonSTCON is NL-complete is actually not so easy. Even showing that nonSTCON is in NL is not so easy; it is equivalent to showing that NL equals coNL. This is a non-trivial result due to Immerman and Szelepcsényi.