I proved that a $\Delta_0$ formula $\varphi$ cannot be independent of ZFC, i.e.,
(*) If $\varphi$ is a $\Delta_0$ formula and $V$ is a model of ZFC, then one of the followings must hold:
(i) $V\models\varphi$, or
(ii) $V\models\neg\varphi$
I proved (*) by a Boolean valued model $V^B$:
Lemma
(i) If $\varphi(x,\ldots)$ is a $\Delta_0$ formula then $\lVert\varphi(\check{x},\ldots)\rVert\in\{0,1\}$.
(ii) If $\varphi$ is a $\Delta_0$ formula, then $\varphi(x,\ldots)$ if and only if $\lVert\varphi(\check{x},\ldots)\rVert=1$.
But I'm still not sure whether the proposition (*) is true since it seems powerful. If it is true, is there a proof without utilizing the Boolean valued model?
There are several issues here.
The first is that formulas are generally with free variables. So what do you mean when you say that a formula is independent of a theory? Do you mean its universal closure (i.e. add $\forall$ on all the free variables?)
Or do you mean that if two models of set theory agree on elementary diagram of some object(s) $x$, then they also agree on their $\Delta_0$ properties?
The second is that you only proved that logic is Boolean. If $V$ is a structure of a language $\cal L$ (in our case, the $\{\in\}$), then for every sentence, $\varphi$, either $V\models\varphi$ or $V\models\lnot\varphi$. This is not about $\Delta_0$ or anything else. Just about how truth is defined—relative to a given structure—for every sentence.
But since you use Boolean valued models, I'm guessing that you're trying to get something out of forcing.
If $V$ is a model of $\sf ZF$, then in any generic extension $V[G]$, we have that $V$ is a transitive class of $V[G]$ (in the broad sense, at least, since it might not be definable in the absence of choice in $V[G]$). And since $\Delta_0$ formulas are absolute between transitive classes that contain all the relevant sets (i.e. the parameters), you immediately get that every $\Delta_0$ formula is true in $V$ if and only if it is true in $V[G]$.
The same can be said about the construction of inner models, e.g. $L$ or $\rm HOD$.
In fact, more can be said: if $W$ is an end-extension of $V$, then they also agree on $\Delta_0$ truths with respect to sets in $V$. (Here an end-extension means that we add new ordinals to $V$, but there is no $x\in W\setminus V$ whose rank is an ordinal in $V$.)
So why does [presumably] Jech even want to prove that sort of lemma? For starters, it's a good exercise in understanding how forcing works. Secondly, it is useful for later proofs. Thirdly, at that point, we haven't talked about the generic extension, where $V$ is a transitive subclass, so we can't quite appeal to that sort of absoluteness.