Decide if for each set of sentences such that $Δ⊨φ$, where $φ$ is sentence, there exists minimal (in sense of $\subseteq$) $Δ′⊆Δ$ satisfying $Δ′⊨φ$?
My answer is: Yes, there exists. Such set is independent. We define it as:
$\forall_{φ∈Δ′} Δ′∖{φ}⊭φ$.
So we can only show that such $\Delta'$ exists (independent set).
Am I right ?
Assuming you are working in ordinary first order logic, then if $\Delta \models \phi$, there exists a subset $\Delta'$ of $\Delta$ such that $\Delta' \models \phi$ and $\Gamma \not\models \phi$ for any proper subset $\Gamma$ of $\Delta'$. However your attempted definition of this minimal $\Delta'$ gives the desired property of $\Delta'$ but doesn't show that it exists.
To see that the required minimal $\Delta'$ does exist, note that if $\Delta \models \phi$, then $\Gamma \models \phi$ for some finite subset $\Gamma = \{\phi_1, \ldots, \phi_n\}$ of $\Delta$. Now you can repeatedly remove from $\Gamma$ the first $\phi_i$ such that $\Gamma - \{\phi_i\} \models \phi$ until no such $\phi_i$ exist and the resulting set of formulas will be a minimal set of the sort you are looking for.
However, $\Delta'$ is not uniquely determined. E.g., take $\phi$ to be $a \neq a$ and $\Delta = \{x = y, y = z, x \neq z, u = v, v = w, u \neq w\}$. Then you could take $\Delta'$ to be either of $\{x = y, y = z, x \neq z\}$ or $\{u = v, v = w, u \neq w\}$.