[proofplan]
We prove first that a sequential inductive limit of finite-dimensional $C^*$-algebras has the finite-dimensional approximation property, using the density of the canonical union of the stage images. Conversely, starting from a separable AF algebra, we choose a countable dense sequence and build an increasing sequence of finite-dimensional subalgebras that approximates longer and longer initial segments of that sequence. The only non-formal ingredient is the standard perturbation stability of finite-dimensional $C^*$-subalgebras, equivalently the stability of matrix-unit relations. The closure of the increasing union is then all of $A$, and the universal property of the $C^*$-inductive limit identifies this closure with the inductive limit of the constructed sequence.
[/proofplan]
[step:Approximate finite sets inside one stage of a finite-dimensional inductive limit]
Assume that $A$ is $*$-isomorphic to a $C^*$-inductive limit of a sequence of finite-dimensional $C^*$-algebras. Replacing $A$ by an isomorphic copy, write
\begin{align*}
A=\varinjlim (A_n,\varphi_n)
\end{align*}
with canonical $*$-homomorphisms
\begin{align*}
\psi_n:A_n\to A.
\end{align*}
For $m\leq n$, let $\varphi_{m,n}:A_m\to A_n$ denote the composite connecting map, with $\varphi_{n,n}:=\operatorname{id}_{A_n}$. The canonical maps satisfy
\begin{align*}
\psi_m=\psi_n\circ \varphi_{m,n}.
\end{align*}
By the density conclusion in the existence theorem for $C^*$-inductive limits [citetheorem:8590], applied to the sequential directed system $(A_n,\varphi_{m,n})$, the subset
\begin{align*}
U:=\bigcup_{n=1}^{\infty}\psi_n(A_n)
\end{align*}
is dense in $A$. The hypotheses of [citetheorem:8590] are satisfied because $\mathbb N$ is directed by its usual order, each $A_n$ is a $C^*$-algebra, each composite map $\varphi_{m,n}:A_m\to A_n$ is a $*$-homomorphism, $\varphi_{n,n}=\operatorname{id}_{A_n}$, and the composites satisfy $\varphi_{n,k}\circ\varphi_{m,n}=\varphi_{m,k}$ whenever $m\leq n\leq k$.
Let $F\subset A$ be finite and let $\varepsilon>0$. For each $a\in F$, choose an element $u_a\in U$ such that
\begin{align*}
\|a-u_a\|_A<\varepsilon.
\end{align*}
Since $F$ is finite, there are indices $n(a)\in\mathbb N$ and elements $b_a\in A_{n(a)}$ such that
\begin{align*}
u_a=\psi_{n(a)}(b_a)
\end{align*}
for every $a\in F$. Choose $N\in\mathbb N$ with $N\geq n(a)$ for every $a\in F$. Then
\begin{align*}
u_a=\psi_N(\varphi_{n(a),N}(b_a))\in \psi_N(A_N)
\end{align*}
for every $a\in F$.
The image
\begin{align*}
E:=\psi_N(A_N)
\end{align*}
is a finite-dimensional $C^*$-subalgebra of $A$, because $A_N$ is finite-dimensional and $\psi_N$ is a $*$-homomorphism. Hence
\begin{align*}
\operatorname{dist}(a,E)\leq \|a-u_a\|_A<\varepsilon
\end{align*}
for every $a\in F$. Thus $A$ is AF.
[guided]
The goal is to show that the inductive-limit description implies the finite-dimensional approximation property. Assume, after composing with the given $*$-isomorphism if necessary, that
\begin{align*}
A=\varinjlim (A_n,\varphi_n),
\end{align*}
where every $A_n$ is finite-dimensional. Let
\begin{align*}
\psi_n:A_n\to A
\end{align*}
be the canonical $*$-homomorphism from the $n$th stage into the limit. For $m\leq n$, define
\begin{align*}
\varphi_{m,n}:A_m\to A_n
\end{align*}
to be the composite connecting map, and set $\varphi_{n,n}:=\operatorname{id}_{A_n}$. The compatibility condition for the limit says that
\begin{align*}
\psi_m=\psi_n\circ \varphi_{m,n}.
\end{align*}
The key structural fact about a $C^*$-inductive limit is that the stage images have dense union. We invoke the density conclusion in the existence theorem for $C^*$-inductive limits [citetheorem:8590]. Its hypotheses are met here: $\mathbb N$ is directed by its usual order, each $A_n$ is a $C^*$-algebra, each map $\varphi_{m,n}:A_m\to A_n$ is a $*$-homomorphism, $\varphi_{n,n}=\operatorname{id}_{A_n}$, and the compatibility identity $\varphi_{n,k}\circ\varphi_{m,n}=\varphi_{m,k}$ holds for $m\leq n\leq k$ by definition of the composite connecting maps. Therefore the set
\begin{align*}
U:=\bigcup_{n=1}^{\infty}\psi_n(A_n)
\end{align*}
is dense in $A$.
Now fix a finite subset $F\subset A$ and a tolerance $\varepsilon>0$. Since $U$ is dense in $A$, for each $a\in F$ we may choose $u_a\in U$ satisfying
\begin{align*}
\|a-u_a\|_A<\varepsilon.
\end{align*}
Each chosen $u_a$ lies in some stage image, so there are an index $n(a)\in\mathbb N$ and an element $b_a\in A_{n(a)}$ such that
\begin{align*}
u_a=\psi_{n(a)}(b_a).
\end{align*}
Because $F$ is finite, the finitely many indices $n(a)$ have a common upper bound. Choose $N\in\mathbb N$ such that $N\geq n(a)$ for every $a\in F$. By the compatibility relation for the canonical maps,
\begin{align*}
u_a=\psi_N(\varphi_{n(a),N}(b_a)).
\end{align*}
Thus every approximant $u_a$ belongs to the single stage image $\psi_N(A_N)$.
Define
\begin{align*}
E:=\psi_N(A_N).
\end{align*}
This is a finite-dimensional $C^*$-subalgebra of $A$: it is the image of a finite-dimensional $C^*$-algebra under a $*$-homomorphism, hence a finite-dimensional $*$-subalgebra and therefore closed. Since each $u_a$ lies in $E$, we have
\begin{align*}
\operatorname{dist}(a,E)\leq \|a-u_a\|_A<\varepsilon
\end{align*}
for every $a\in F$. This is exactly the AF approximation property.
[/guided]
[/step]
[step:Choose a dense sequence and state the perturbation tool]
Assume conversely that $A$ is separable and AF. Since $A$ is separable, choose a sequence
\begin{align*}
(a_n)_{n\in\mathbb N}
\end{align*}
in $A$ whose range is dense in $A$.
We use the following precise perturbation input, the finite-dimensional enlargement theorem for AF algebras. Let $C\subset A$ be a finite-dimensional $C^*$-subalgebra, let $F\subset A$ be finite, and let $\eta>0$. Then there exist a finite subset $G\subset A$ and a number $\delta>0$ such that if $B\subset A$ is a finite-dimensional $C^*$-subalgebra satisfying
\begin{align*}
\operatorname{dist}(x,B)<\delta
\end{align*}
for every $x\in F\cup G$, then there exists a finite-dimensional $C^*$-subalgebra $D\subset A$ with $C\subset D$ and
\begin{align*}
\operatorname{dist}(x,D)<\eta
\end{align*}
for every $x\in F$. This is the standard perturbation stability theorem for finite-dimensional $C^*$-subalgebras, equivalently the stability of finite systems of matrix-unit relations.
[guided]
The separability hypothesis is used only to reduce the approximation problem to a sequence of finite sets. Since $A$ is separable, there is a sequence
\begin{align*}
(a_n)_{n\in\mathbb N}
\end{align*}
in $A$ whose range is dense in $A$.
The converse direction also needs a way to keep finite-dimensional approximants nested. The AF hypothesis by itself gives, for each finite set, some finite-dimensional subalgebra close to that set, but it need not contain the finite-dimensional algebra chosen at the previous stage. The finite-dimensional enlargement theorem supplies exactly this missing nesting step. Given a finite-dimensional $C^*$-subalgebra $C\subset A$, a finite subset $F\subset A$, and $\eta>0$, it produces a finite set $G\subset A$ and a number $\delta>0$ such that every finite-dimensional $C^*$-subalgebra $B\subset A$ satisfying
\begin{align*}
\operatorname{dist}(x,B)<\delta
\end{align*}
for every $x\in F\cup G$ can be enlarged, after perturbing the relevant finite-dimensional data, to a finite-dimensional $C^*$-subalgebra $D\subset A$ with $C\subset D$ and
\begin{align*}
\operatorname{dist}(x,D)<\eta
\end{align*}
for every $x\in F$. This is the standard perturbation stability theorem for finite-dimensional $C^*$-subalgebras, equivalently stability of finite systems of matrix-unit relations.
[/guided]
[/step]
[step:Construct nested finite-dimensional subalgebras approximating the dense sequence]
We construct finite-dimensional $C^*$-subalgebras
\begin{align*}
C_1\subset C_2\subset C_3\subset \cdots \subset A
\end{align*}
such that
\begin{align*}
\operatorname{dist}(a_j,C_n)<\frac{1}{n}
\end{align*}
whenever $1\leq j\leq n$.
For $n=1$, the AF property applied to the finite set $\{a_1\}$ and the tolerance $1$ gives a finite-dimensional $C^*$-subalgebra $C_1\subset A$ with
\begin{align*}
\operatorname{dist}(a_1,C_1)<1.
\end{align*}
Assume that $C_{n-1}$ has been constructed for some $n\geq 2$. Set
\begin{align*}
F_n:=\{a_1,\dots,a_n\}.
\end{align*}
Apply the perturbation lemma to the finite-dimensional subalgebra $C_{n-1}\subset A$, the finite set $F_n$, and the tolerance $1/n$. This gives a finite subset $G_n\subset A$ and a number $\delta_n>0$ with the property stated above. Since $A$ is AF, there exists a finite-dimensional $C^*$-subalgebra $B_n\subset A$ such that
\begin{align*}
\operatorname{dist}(x,B_n)<\delta_n
\end{align*}
for every $x\in F_n\cup G_n$. The perturbation lemma then gives a finite-dimensional $C^*$-subalgebra $C_n\subset A$ such that
\begin{align*}
C_{n-1}\subset C_n
\end{align*}
and
\begin{align*}
\operatorname{dist}(a_j,C_n)<\frac{1}{n}
\end{align*}
for every $1\leq j\leq n$. This completes the induction.
[/step]
[step:Show that the increasing union is dense in $A$]
Define
\begin{align*}
C_\infty:=\overline{\bigcup_{n=1}^{\infty}C_n}^{\|\cdot\|_A}.
\end{align*}
Then $C_\infty$ is a closed $C^*$-subalgebra of $A$, because the union is an increasing $*$-subalgebra and the operations and involution are norm-continuous.
We show that $C_\infty=A$. Let $a\in A$ and let $\varepsilon>0$. Since the range of $(a_n)_{n\in\mathbb N}$ is dense in $A$, choose $j\in\mathbb N$ such that
\begin{align*}
\|a-a_j\|_A<\frac{\varepsilon}{2}.
\end{align*}
Choose $n\in\mathbb N$ such that $n\geq j$ and
\begin{align*}
\frac{1}{n}<\frac{\varepsilon}{2}.
\end{align*}
By construction,
\begin{align*}
\operatorname{dist}(a_j,C_n)<\frac{1}{n}.
\end{align*}
Since $C_n\subset C_\infty$, this implies
\begin{align*}
\operatorname{dist}(a_j,C_\infty)\leq \operatorname{dist}(a_j,C_n)<\frac{\varepsilon}{2}.
\end{align*}
Therefore
\begin{align*}
\operatorname{dist}(a,C_\infty)\leq \|a-a_j\|_A+\operatorname{dist}(a_j,C_\infty)<\varepsilon.
\end{align*}
Because $C_\infty$ is closed and every $a\in A$ has distance $0$ from $C_\infty$, we obtain $A=C_\infty$.
[/step]
[step:Identify $A$ with the inductive limit of the nested sequence]
For each $n\in\mathbb N$, define the connecting $*$-homomorphism
\begin{align*}
\iota_n:C_n\to C_{n+1}
\end{align*}
to be the inclusion map. Since each $C_n$ is finite-dimensional, the system
\begin{align*}
(C_n,\iota_n)_{n\in\mathbb N}
\end{align*}
is a sequential inductive system of finite-dimensional $C^*$-algebras. By the construction of the $C^*$-inductive limit and the density of the canonical union, its inductive limit is canonically $*$-isomorphic to the norm closure of
\begin{align*}
\bigcup_{n=1}^{\infty}C_n
\end{align*}
inside $A$. The previous step showed that this closure is $A$. Hence
\begin{align*}
A\cong \varinjlim (C_n,\iota_n)
\end{align*}
as $C^*$-algebras.
Combining this with the first direction proves that a separable complex $C^*$-algebra is AF if and only if it is $*$-isomorphic to the $C^*$-inductive limit of a sequence of finite-dimensional complex $C^*$-algebras.
[/step]
