Let $A_{i}$ be sequence of C* algebras, $\alpha_{i}:A_{i}\rightarrow A_{i+1}$ be sequence of *-homomorphisms. Consider some sub-algebra $B$ of its inductive limit $A$. Denote the canonical homomorphism be $\beta_{i}:A_{i}\rightarrow A$. Are $\beta_{i}^{-1}(B)$ a sub-algebra of $A_{i}$?
I know that the kernal of $\beta_{i}$ form an ideal in $A_{i}$, but I don't know if it works on preimage of other elements.
Yes, and this has nothing to do with inductive limits. If $\alpha\colon A_1\to A_2$ is an algebra homomorphism and $B$ a subalgebra of $A_2$, then $\alpha^{-1}(B)$ is a subalgebra of $A_1$. I assume you know this result for vector spaces and linear maps, so I'll just prove that $\alpha^{-1}(B)$ is closed under taking products.
If $x,y\in \alpha^{-1}(B)$, then $\alpha(x),\alpha(y)\in B$ by definition and therefore $\alpha(xy)=\alpha(x)\alpha(y)\in B$ since $B$ is an algebra. Thus $xy\in \alpha^{-1}(B)$.
The same holds for (unital) $\ast$-homomorphisms and (unital) $\ast$-subalgebras by a similar argument.