[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.[/step]

[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: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.[/step]

[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: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]