Show that there exists Δ0⊆Δ\Delta _0 \subseteq \Delta such that Δ0⊨Δ\Delta _0\models\Delta

Δ\Delta is finite set of sentences over finite structure Σ\Sigma.
Show that there exists Δ0⊆Δ\Delta _0 \subseteq \Delta such that Δ0⊨Δ\Delta _0\models\Delta and moreover sentences in Δ0\Delta_0 are independent, which means that ∀ϕ∈Δ0Δ0∖{ϕ}⊭Δ0\forall_{\phi\in\Delta_0}\Delta_0\setminus\{\phi\}\not\models\Delta_0.

My approach is:
Lets start from Δ0=Δ\Delta_0=\Delta. Then Δ⊨Δ\Delta\models\Delta. If Δ\Delta is independent then proof is finished. However, if Δ\Delta is not independent then there exists ϕ∈Δ0\phi\in\Delta_0 such that Δ0∖{ϕ}⊨Δ0\Delta_0\setminus\{\phi\}\models\Delta_0. Because Δ0=Δ\Delta_0=\Delta we know that Δ0∖{ϕ}⊨Δ\Delta_0\setminus\{\phi\}\models\Delta. Hence, we can: Δ0:=Δ0∖{ϕ}\Delta_0 := \Delta_0\setminus\{\phi\}. Due to the fact that Δ\Delta is finite we can repeat this procedure until Δ0\Delta_0 is dependent.

What about this proof ? Does it work ?

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