This question is similar to my question here, but not the same question.
Let $(A_{i},\alpha_{i})$ be directed system of C* algebras and *-homomorphisms. Let the $\beta_{i}:A_{i}\rightarrow A$ are the canonical *-homomorphisms. Consider some subalgebra $B$ belongs to direct limit $A$. Does it always exists a $j\in I$ such that $B\subseteq \beta_{j}(A_{j})$?
My guess is that is true. Since $A$ is quotient algebra of disjoint union of algebras, we can write every elements $x\in A$ in the form $x=(x_{1},x_{2},...)+N$, with $x_{i}\in A_{i}$, $N$ is the ideal form by equivalence relation. If $B$ is a sub-algebra, then every component of $B$ form a sub-algebra. Is my thought correct?
The inductive limit of C$^*$-algebras is often not the union of the ranges of the $\beta _j$.
For instance, consider the C$^*$-algebra $K$, formed by all compact operators on $\ell ^2$. Also, for each $n$, consider the subset $K_n\subseteq K$, formed by all operators whose matrix $(a_{i, j})_{i, j}$, relative to the canonical basis of $\ell ^2$, have nonzero entries only in the top left $n\times n$ block.
Then
each $K_n$ is a closed $^*$-subalgebra of $K$ (isomorphic to $M_n(\mathbb C)$),
$K_n\subseteq K_{n+1}$, and
the union $\bigcup_nK_n$ is dense in $K$.
With this much information you are able to deduce that $K$ is the inductive limit of the $K_n$, with the connecting maps being the inclusions $K_n\hookrightarrow K_{n+1}$.
Observing that every operator in $\bigcup_nK_n$ has finite rank, we see that $\bigcup_nK_n$ is not equal to $K$. It is only a dense $^*$-subalgebra (and hence can't be the inductive limit in the category of C$^*$-algebras since we want the inductive limit to be a C$^*$-algebra).
Any infinite rank compact operator will therefore generate a subalgebra that is not contained in the union of the ranges of the $\beta _j$.