Use the injective <group homomorphism> $F_2\hookrightarrow\operatorname{SL}_2(\mathbb Z)$ from part (a). If $1\ne g\in F_2$, its image is an integral matrix $M\ne I$. Choose a <prime number> $p$ that does not divide one nonzero entry of $M-I$. The <reduction modulo a prime in an integral matrix group> homomorphism
$$
\operatorname{SL}_2(\mathbb Z)\longrightarrow
\operatorname{SL}_2(\mathbb F_p)
$$
then sends $M$ to a nonidentity element. Its target is a <finite group>, so the composite map separates $g$ from the identity. Hence $F_2$ is a <residually finite group>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 133 / 1 / b / Solution

Every finitely generated <free group> is a <residually finite group>. For $F_2$, the <ping-pong lemma> embeds it into $\operatorname{SL}_2(\mathbb Z)$, and <reduction modulo a prime in an integral matrix group> separates each nonidentity element in a finite quotient.

Residual finiteness of a free group

For a <prime number> $p$, reducing every entry modulo $p$ defines a <group homomorphism> from an integral matrix group to the corresponding matrix group over $\mathbb F_p$. The target is finite. If an integral matrix $M$ is not the identity, choosing $p$ not to divide one nonzero entry of $M-I$ ensures that its reduction is also nonidentity.

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 Incoming links: Reduction modulo a prime in an integral matrix group