If $L$ is a regular language then so is $L/a =\{w | wa ∈ L\}$, where $L$ is a language over $\Sigma$ and $a \in \Sigma$
This proof uses both construction and knowledge of closure properties of regular languages.

First, we know that regularity of languages is closed under the reversal operation, see proof of Thm. 4.2 here.

We can then define an operation named chop such that:

$$
\text{chop}(L)=\{w|aw \in L\}, \text{ where } a \in \Sigma, 
$$

We can construct an NFA for chop, which simply omits the first transition (hence removing the first character):



This NFA is incomplete, but is just for the sake of illustration

Therefore the regularity is closed under the chop operation.

Finally, observe that we can produce R/a operation using chop and reversal:

$$
L/a = (\text{chop}(L^{R}))^{R}
$$

Hence L/a has been constructed using closure properties so must be regular.

Hope this helps!

Source: Formal Languages module, second year, BSc Computer Science University of Warwick.

Edit: capitalisation of first word.