If we are talking about Hilbert-style proof system, this rule is not an inference rule within the theory. A proof of "if $\Gamma \vdash P(c)$ then $\Gamma \vdash (\forall x)P(x)$" would be a meta-proof, i.e. a proof of the sentence about the first-order theory, but not a sentence in the theory.
So an informal proof is as follows. Since $\Gamma \vdash P(c)$, there exists a proof, i.e. a finite sequence of sentences each of which is either an axiom, or a sentence of $\Gamma$, or a sentence derived by inference rule of the theory. Now we pick a new variabe $y$ such that $y$ does not appear in the proof. This is possible because (1) the proof must be of a finite sequence, hence contains a finite number of variables; (2) in a first-order theory, there are infinitely (though countable) many variables. After we pick such a $y$, we replace $c$ by $y$ in the proof, yielding a new finite sequence of sentences. It is not difficult to verify that this new sequence is indeed a proof in the theory. That is, one has to verify that every sentence in this new sequence is either an axiom, or a sentence of $\Gamma$, or dervied by inference rules. Note $\Gamma$ does not contain the constant symbol $c$. So the above verification is possible. Finally, in this new proof, the last sentence must be $P(y)$. Now applying Gen rule, we can obtain $(\forall y)P(y)$. Deriving $(\forall x)P(x)$ from $(\forall y)P(y)$ is straightforward.
The above meta-proof is "informal" because it is stated in English. I think it is possible to formulate the above proof in a first-order set theory, although I'm not certeain. Nevertheless, I think it suffices to accept such an informal proof, just like other meta-proofs and meta-statements that discuss various properties about the subject-matter first-order theory, such as models, consistency, completeness, etc.