Here is another proof that I like, although it is not as simple as modding out by a cofinite ideal as per @Carl-FredrikNybergBrodda's answer. This approach I learned from Stalling's old geometric group theory lecture notes. Hopefully I am correctly recalling it. Since all free monoids on a countable generating set embed in the free monoid on 2 generators $\mathbf 0$, $\mathbf 1$ it is enough to do that case.

I'll give a representation of the free monoid on $2$ generated by $2\times 2$ nonnegative invertible matrices that is particularly easy to verify is faithful (but this representation does not extend faithfully to the free group). Since the group of $2\times 2$ invertible integer matrices is clearly residually finite by taking the entries modulo large primes, we are done.

Map $\mathbf 0$ to $\begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}$ and map $\mathbf 1$ to $\begin{bmatrix} 1 & 1\\ 0 & 2\end{bmatrix}$. Then a simple induction shows a word $w$ maps to $\begin{bmatrix} 1 & d(w)\\ 0 & 2^{|w|}\end{bmatrix}$ where $|w|$ is the length of $w$ and $d(w)$ is the natural number whose base $2$ expansion is $w$ (with perhaps some leading zeroes). Knowing the length of $w$ and the integer $d(w)$ lets you recover $w$ since from the length you know how many leading zeroes to add.

Here is another proof that I like, although it is not as simple as modding out by a cofinite ideal as per @Carl-FredrikNybergBrodda's answer. This approach I learned from Stalling's old geometric group theory lecture notes. Hopefully I am correctly recalling it. Since all free monoids on a countable generating set embed in the free monoid on 2 generators $\mathbf 0$, $\mathbf 1$ it is enough to do that case.

I'll give a representation of the free monoid on $2$ generated by $2\times 2$ nonnegative matrices (invertible over $\mathbb Z[1/2]$) that is particularly easy to verify is faithful (but this representation does not extend faithfully to the free group). Since the monoid of $2\times 2$ integer matrices is clearly residually finite by taking the entries modulo large primes, we are done.

Map $\mathbf 0$ to $\begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}$ and map $\mathbf 1$ to $\begin{bmatrix} 1 & 1\\ 0 & 2\end{bmatrix}$. Then a simple induction shows a word $w$ maps to $\begin{bmatrix} 1 & d(w)\\ 0 & 2^{|w|}\end{bmatrix}$ where $|w|$ is the length of $w$ and $d(w)$ is the natural number whose base $2$ expansion is $w$ (with perhaps some leading zeroes). Knowing the length of $w$ and the integer $d(w)$ lets you recover $w$ since from the length you know how many leading zeroes to add.