This is by far not a complete solution, but at least a partial one, plus a potentially useful interpretation and, perhaps, a direction to pursue.
The way I understand the problem, given a $K$-element set $X$, we call its $k$-element subsets blocks, and we say that a block $B\subset X$ separates the elements $x_1,x_2\in X$ if exactly one of these elements is contained in $B$. We want to estimate the smallest possible number of blocks, say $n=n(K,k)$, separating every pair of elements of $X$. We can in fact confine to the range $k\le K/2$ since $n(K,K-k)=n(K,k)$, as it follows by looking at the complements of the blocks.
For $k=1$, one needs at least $K-1$ blocks (otherwise there will be at least two elements not contained in any block, hence not separated by any block). Considering, on the other hand, the system of $K-1$ pairwise disjoint, one-element blocks, we get
$$ n(K,1) = K-1. $$
For $k=2$, one can identify $X$ with the set of vertices, and blocks with the edges of a graph. Two vertices $x_1,x_2\in X$ are separated whenever at least one of them is adjacent to a vertex other than another one. Thus, we want to minimize the number of edges subject to the condition that the graph does not have isolated edges, and has at most one isolated vertex; in other words, every connected component of the graph has at least three vertices, except that there can be one component containing an isolated vertex. In the optimal case, every component is a tree (take a spanning tree and remove all other edges), and then the number of edges of the graph is $K$ less the number of components. Thus, minimizing the number of edges is equivalent to maximizing the number of components. Consequently, all components must be paths of length $2$, with an obvious adjustment in one component if $K$ is not divisible by $3$. This gives
$$ n(K,2) = \begin{cases}
2K/3 \ &\text{if}\ K\equiv 0\pmod 3, \\
2(K-1)/3 &\text{if}\ K\equiv 1\pmod 3, \\
2(K-2)/3 + 1 &\text{if}\ K\equiv 2\pmod 3.
\end{cases} $$
In the general case, the following approach can be useful. Identifying subsets of $X$ with the vectors of $\mathbb F_2^K$, consider the bipartite graph on the vertex set $V_k\cup V_2$, where $V_k$ is the set of all vectors of weight $k$, and $V_2$ is the set of all vectors of weight $2$, with $b\in V_k$ adjacent to $x\in V_2$ if and only if $b$ is not orthogonal to $x$. For any $x_1,x_2\in X$ with $x_1\ne x_2$, the vector in $V_2$ corresponding to the subset $\{x_1,x_2\}\subset X$ is adjacent to a vector $b\in V_k$ corresponding to the block $B\subset X$ if and only if $x_1$ and $x_2$ are separated by $B$. It follows that $n(K,k)$ is the smallest possible number of vertices from $V_k$ dominating any vertex from $V_2$.
In principle, you can now take any of the numerous estimates for the domination number of a graph and try to adapt it to the settings described. The simplest estimate is as follows: since any $b\in V_k$ dominates $k(K-k)$ vertices from $V_2$, to dominate all $\binom K2$ vertices of $V_2$ one needs at least $K(K-1)/(2k(K-k))$ vertices $b\in V_k$; that is,
$$ n(K,k) \ge \frac{K(K-1)}{2k(K-k)}. $$
