The Leray spectral sequence for $\mathbb{A}^{1}_{k} \to \operatorname{Spec} k$ is \begin{align} \mathrm{E}_{2}^{p,q} = \mathrm{H}^{p}(\Gamma,\mathrm{H}_{et}^{q}(\mathbb{A}^{1}_{k^{\mathrm{sep}}},\mathbb{G}_{m})) \implies \mathrm{H}_{et}^{p+q}(\mathbb{A}^{1}_{k},\mathbb{G}_{m}) \end{align} with differentials $\mathrm{E}_{2}^{p,q} \to \mathrm{E}_{2}^{p+2,q-1}$. Since $\operatorname{Pic}(\mathbb{A}^{1}_{k^{\mathrm{sep}}}) = 0$ and $\operatorname{Pic}(\mathbb{A}^{1}_{k^{\mathrm{sep}}} \setminus \{0\}) = 0$, this and the analogous spectral sequence for $\mathbb{A}^{1}_{k} \setminus \{0\} \to \operatorname{Spec} k$ gives a commutative diagram
$\require{AMScd}$
\begin{CD}
    0 @>>> \operatorname{Br}(k) @>\xi_{1}>>  \operatorname{Br}(\mathbb{A}^{1}_{k}) @>\xi_{2}>> \mathrm{H}^{0}(\Gamma,\operatorname{Br}(\mathbb{A}^{1}_{k^{\mathrm{sep}}})) \\
    @. @V\rho_{1}VV @V\rho_{2}VV @V\rho_{3}VV \\
    0 @>>> \operatorname{Br}(k) \oplus \mathrm{H}^{2}(\Gamma,\mathbb{Z}) @>>\xi_{1}'>  \operatorname{Br}(\mathbb{A}^{1}_{k} \setminus \{0\}) @>>\xi_{2}'> \mathrm{H}^{0}(\Gamma,\operatorname{Br}(\mathbb{A}^{1}_{k^{\mathrm{sep}}} \setminus \{0\})) \\
\end{CD}
with exact rows, where the $\rho_{i}$ are induced by restriction along the open immersion $\mathbb{A}^{1}_{k} \setminus \{0\} \subset \mathbb{A}^{1}_{k}$.

Since $\mathbb{A}_{k}^{1}$ and $\mathbb{A}_{k^{\mathrm{sep}}}^{1}$ are regular Noetherian (and also since $\mathrm{H}^{0}(\Gamma,-)$ is left exact), the restriction maps $\rho_{2}$ (and hence also $\rho_{1})$ and $\rho_{3}$ are injective. Suppose $\alpha_{1}' \in \mathrm{H}^{2}(\Gamma,\mathbb{Z})$ is such that $\xi_{1}'(\alpha_{1}') = \rho_{2}(\alpha_{2})$ for some $\alpha_{2} \in \operatorname{Br}(\mathbb{A}^{1}_{k})$. Then $\rho_{3}(\xi_{2}(\alpha_{2})) = \xi_{2}'(\rho_{2}(\alpha_{2})) = \xi_{2}'(\xi_{1}'(\alpha_{1}')) = 0$, thus $\xi_{2}(\alpha_{2}) = 0$, thus there exists some $\alpha_{1} \in \operatorname{Br}(k)$ such that $\xi_{1}(\alpha_{1}) = \alpha_{2}$, but injectivity of $\xi_{1}'$ implies that $\alpha_{1} = 0$ and $\alpha_{1}' = 0$.