Direct proof
We want to prove that a channel given in the form
$$\Phi(\rho) = \sum_a A_a \rho A_a^\dagger\tag1$$
can equivalently be represented as a unitary evolution in an enlarged space. I'll actually consider the more general case of an arbitrary (CP) map of the form (1) (that is, I won't assume the normalisation $\sum_a A_a^\dagger A_a=I$).
A direct way to do this is to define the operator $V$ with action
$$V|\psi\rangle\equiv \sum_a (A_a|\psi\rangle)\otimes |a\rangle.$$
If $\Phi$ is a channel, that is, $\sum_a A_a^\dagger A_a=I$, then $V$ is an isometry. The vectors $|a\rangle$ are here an (arbitrary) orthonormal basis in an auxiliary space. The auxiliary space needs to have a dimension equal to the number of Kraus operators for this expression to work. You could have a larger ancillary space if you're fine with $V$ being a partial isometry instead, but you don't really gain anything doing this.
With this $V$, we can see that the map can be equivalently written as
$$\Phi(\rho) = \operatorname{Tr}_2[V\rho V^\dagger].\tag2$$
This is the so-called Stinespring representation of the map/channel. We can then also equivalently write (2) as
$$\Phi(\rho) = \operatorname{Tr}_2[U(\rho\otimes |u\rangle\!\langle u|)U^\dagger)],$$
for any pure state $|u\rangle$ living in the auxiliary space, defining $U$ as such that $U(|i\rangle\otimes|u\rangle)=V|i\rangle$.
This can always be done, and if $\Phi$ is a channel, then $V$ is an isometry, and $U$ is a unitary.
Proof with explicit componentwise expressions
From the Kraus representation, $\Phi(\rho)=\sum_a A^a \rho A^{a\dagger}$, making the indices explicit, we get
$$\Phi(\rho)_{ij}=\sum_{a,k,\ell}A^a_{ik}A^{a*}_{j\ell}\rho_{k\ell}.\tag1$$
On the other hand, unravelling the second expression we have for a generic $\sigma$ (let me here use numbers instead of latin letters for the indices, for better clarity, as well as Einstein's notation for repeated indices),
$$[U(\rho \otimes \sigma) U^\dagger]_{1234}=U_{1256}U^{*}_{3478}\rho_{57}\sigma_{68}.$$
Note that here the first two indices, ($1$ and $2$) correspond to the "output space" of the operator $\Phi(\rho)$, while the other two ($3$ and $4$) correspond to its "input space". Similarly, $2$ and $4$ live in the second Hilbert space, while $1$ and $3$ live in the first one.
Tracing with respect to the second Hilbert space amounts to introducing a $\delta_{24}$ factor, and we thus get
$$\left\{\mathrm{Tr}_B\left[ U(\rho \otimes \sigma) U^\dagger \right]\right\}_{13}
=U_{1256}U^{*}_{3478}\rho_{57}\sigma_{68} \color{red}{\delta_{24}}
=U_{1256}U^{*}_{3278}\rho_{57}\sigma_{68}.$$
If we take $\sigma$ to be a pure state, for example $\sigma=\lvert0\rangle\!\langle0\rvert$, so that $\sigma_{68}=\delta_{60}\delta_{80}$, then we have
$$\left\{\mathrm{Tr}_B\left[ U(\rho \otimes \lvert0\rangle\!\langle0\rvert) U^\dagger \right]\right\}_{13}
=U_{1250}U^*_{3270}\rho_{57}.$$
Going back to using the standard notation for the indices, and making explicit the sums, we have
$$\left\{\mathrm{Tr}_B\left[ U(\rho \otimes \lvert0\rangle\!\langle0\rvert) U^\dagger \right]\right\}_{ij}
=\sum_{a,k,\ell}U_{iak0}U^*_{ja\ell0}\rho_{k\ell}.\tag2$$
This expression is equivalent to (1), defining $A^a_{ik}\equiv U_{iak0}$ and $A_{j\ell}^{a*}\equiv U^*_{ja\ell0}$.
