= Solution
We use the <Van den Berg-Kesten inequality>: for increasing events $A,B$ under independent <bond percolation>, their disjoint occurrence $A\mathbin\square B$, meaning that they have disjoint sets of open <edges> witnessing the two events, satisfies $\mathbb P_p(A\mathbin\square B)\le\mathbb P_p(A)\mathbb P_p(B)$. The finite-edge statement applies here; connection to a finite-box boundary can always be witnessed before the first exit from that box.
On the event $0\leftrightarrow\partial\Lambda_{m+n}$ choose a simple open path to that boundary, and let $y$ be its first point on $\partial\Lambda_m$. Its initial segment witnesses $0\leftrightarrow\partial\Lambda_m$ through $y$, inside $\Lambda_m$. The remaining segment eventually reaches maximum-norm distance at least $n$ from $y$, because its endpoint has norm $m+n$ while $\|y\|_\infty=m$. Stop that segment when it first reaches $y+\partial\Lambda_n$. These two segments use disjoint <edges>.
For fixed $y$, let $A_y$ be connection from zero to $y$ inside $\Lambda_m$, and $B_y$ be connection from $y$ to $y+\partial\Lambda_n$ inside $y+\Lambda_n$. Then $\mathbb P_p(A_y)\le\beta_m$ and, by translation invariance, $\mathbb P_p(B_y)=\beta_n$. The <union bound> and the stated disjoint-occurrence inequality give
$$
\boxed{\beta_{m+n}\le\sum_{y\in\partial\Lambda_m}\mathbb P_p(A_y\mathbin\square B_y)\le|\partial\Lambda_m|\beta_m\beta_n.}
$$
Set $x_n=\log\beta_n$ and $\alpha_n=\log|\partial\Lambda_n|$. The previous part applies, since
$$
|\partial\Lambda_n|=(2n+1)^d-(2n-1)^d=O_d(n^{d-1}),\qquad \alpha_n/n\longrightarrow0.
$$
Moreover a straight path of $n$ open bonds has <probability> $p^n$, so $p^n\le\beta_n\le1$. Thus the limit is finite and
$$
\boxed{\gamma=\lim_{n\to\infty}\frac{\log\beta_n}{n}\in[\log p,0],\qquad
\beta_n\ge|\partial\Lambda_n|^{-1}e^{n\gamma}.}
$$
Equivalently, for the nonnegative decay rate $\kappa=-\gamma$, the bound is $\beta_n\ge|\partial\Lambda_n|^{-1}e^{-n\kappa}$.
The printed final exponent has the opposite sign while retaining the definition of $\gamma$ as the limit of $n^{-1}\log\beta_n$. With that definition it cannot be correct in general. Indeed, if $(2d-1)p<1$, count self-avoiding paths of length $\ell\ge n$ to get
$$
\beta_n\le\sum_{\ell\ge n}2d(2d-1)^{\ell-1}p^\ell
=\frac{2d}{2d-1}\frac{((2d-1)p)^n}{1-(2d-1)p}.
$$
Hence $\gamma\le\log((2d-1)p)<0$. A lower bound with $e^{-n\gamma}$ would eventually exceed one, because the boundary factor grows only polynomially. \b[The valid bound is the boxed expression with $e^{n\gamma}$, or the equivalent expression using $\kappa=-\gamma$.]
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