I want to know if it's a good counterexample!
Proof: Let K be a simplicial complex given by the vertices {1,2,3} and the edges {1,2} and {2,3}. That is, K has 2 edges and three vertices arranged in a straight line. Now let's consider the subcomplex L given by the vertex {2}. Clearly, L is a proper subcomplex of K, since it does not contain the edge {1,2}. The complement K-L consists of the vertices {1,3} and the edge {1,3}. However, K-L is not a subcomplex of K, as it contains the edge {2,3}. The statement is false.
The counterexample is good, but, I don't quite follow the details of your argument. For example, $K$ itself does not contain the edge $\{1,3\}$, so how could $K-L$ contain that edge?
What I would say instead is that $K-L$ contains the vertices $\{1\}$ and $\{3\}$, and $K-L$ also contains the interiors of the two edges $\{1,2\}$ and $\{2,3\}$, but $K-L$ does not contain the endpoint $\{2\}$ of the two edges $\{1,2\}$ and $\{2,3\}$. Therefore $K-L$ is not a subcomplex, because if a subcomplex contains (the interior of) a certain simplex $\sigma$ of $K$ then it must also contain every subsimplex of $\sigma$.
In fact this argument can be souped up to prove that if $K$ is a connected simplicial complex and $L \subset K$ is a subcomplex then $K-L$ is a subcomplex if and only if $L= \emptyset$ or $L=K$. More generally, for any simplicial complex $K$ and subcomplex $L \subset K$, $K-L$ is a subcomplex if and only if $L$ can be expressed as the union of a collection of connected components of $K$.