Let $X$ be a compact Hausdorff space. Let $(x_n)$ be a sequence in $X$.
Assume $X$ has no isolated points.
Define $A_n = C(X, M_{2^n}(\mathbb{C}) )$ and define $\phi_{n+1,n} : A_n \to A_{n+1}$ by $$\phi_{n+1,n}(f)=\begin{bmatrix}f&0\\0&f(x_n)\mathbb{1}_{A_n}\end{bmatrix}.$$
Show that $A:= \varinjlim A_n$ is simple if and only if the sequence $(x_n)$ is dense in $X$.
Solution:
First, observe that the connecting maps are injective, so we can think about the direct limit as the closure of the increasing union of the $A_n$'s.
Also,ideals in $A_n$ are of the form $C(X\setminus Y,M_{2^n})$, i.e. functions that vanish at $Y$, where $Y\subseteq X$ is a closed subset.
$\Rightarrow)$ Suppose $(x_n)$ is not dense in $X$. Define for $n\in \mathbb{N}$ $$I_n=\left\{(f_{ij})_{i,j}\in C(X,M_{2^n})|\forall i,j \ \ f_{ij}|_{\overline{\cup_{n}\{x_n\}}}=0\right\}.$$ The sequence is not dense, so $I_n$ is a non-trivial ideal in $A_n$. Define $I=\overline{\cup I_n}$ ideal in $A$, because the union is increasing. We have $I \ne (0)$ because $I_1\ne (0)$ and $I_n\subseteq I_{n+1}$. Also $I \ne A$ , otherwise $1_A \in I$ so there exists some $n$ and $x\in I_n$ s.t. $\|1_A-x\|<1$, thus $x$ is invertible in $A$. But all the $A_n$'s contain $1_A$, and in particular that implies $x$ is invertible in $I_n$, contradicting the fact $I_n\ne A_n$.
$\Leftarrow$) Suppose $I$ is a proper ideal in $A$. $I=\overline{\cup (I\cap A_n)}=\overline{\cup I_n}$, and each $I_n$ correspond to some $Y_n\subseteq X$ closed, as I mentioned before. $I$ is prper, thus any $I_n$ is proper, and $\forall n, \ Y_n \ne \emptyset$.
From the inclusion $I_n\subseteq I_{n+1}$, we know that $\phi (I_n)\subseteq I_{n+1}$.
Consequently, if $f\in I_n$ (vanishes at $Y_n$) then $\begin{bmatrix}f&0\\0&f(x_n)\mathbb{1}_{A_n}\end{bmatrix}|Y_{n+1}=0$, i.e. $Y_{n+1}\subseteq Y_n$ and $x_n\in Y_n$.
Similarily, by applying $\phi _{n+m,n}$, we get $\forall m\geq n \ \ x_m\in Y_n$.$(*)$(This conclusion is right for each $n\in \Bbb{N})$
Assume by contradiction that there exists $k$ s.t. $Y_k \subsetneq X$.
From density of $\{x_n\}_n$ there exists $n$ s.t. $x_n \notin Y_k$.
We've assumed that $X$ has no isolated points, therefore there exist infinitely many points from the sequence $\{x_n\}_n$ which are not in $Y_k$.
This is a contadiction to $(*)$.
So, $\forall k \ \ Y_k=X$ implies $I_k=(0)$ $\rightarrow I=(0)$, as required.
Note that in your definition of $I_n$ you have some confusing use of $n$, which is fixed but also appears in $\cup_n\{x_n\}$, which is probably better written as $\overline{\{x_k:\ k\in\mathbb N\}}$.
I'm not particularly comfortable with the way you argue that $I\ne A$. It is enough to show that $I$ cannot contain any function that is nonzero on $\overline{\{x_n:\ n\in\mathbb N\}}$.
For the converse, It think that your argument with the $Y_k$ is fine.