For a pointed metric space $(M,d,0)$ (meaning $(M,d)$ is a metric space and $0\in M$ is a distinguished point), $\text{Lip}_0(M)$ is the Banach space of all real-valued Lipschitz functions $f:M\to \mathbb{R}$ such that $f(0)=0$. The Lipschitz free space $\mathcal{F}(M)$ is the closure in $\text{Lip}_0(M)^*$ of the space of molecules. A molecules is a functional of the form $\delta_x-\delta_y$, where $\delta_x\in \text{Lip}_0(M)^*$ is the Dirac evaluation (at $x$) functional whose action is given by $\delta_x(f)=f(x)$.
It holds that $\mathcal{F}(M)^*=\text{Lip}_0(M)$. For distinct $x,y\in M$, let $Q(x,y)=\frac{\delta_x-\delta_y}{d(x,y)}\in S_{\mathcal{F}(M)}$. For $x\in M$, let $Q(x,x)=0\in \mathcal{F}(M)$. Here, $S_{\mathcal{F}(M)}$ denotes the closed unit sphere of $\mathcal{F}(M)$.
Does it hold that $$\inf \{\|a-b\|:a,b\in Q(K\times K), a\neq b\}>0?$$
Nope, at least for a separable, uncountable metric space $(M,d,0)$ as your condition would force $F(M)$ to be non-separable (the set of normalised molecules would be uncountable and discrete in the norm), which is not the case for $M$ separable.