The exercise asks to show that $\mathcal{B}(\mathbb{R})\subset \mathcal{B}(\mathbb{C})$.
Is this solution correct?
Let $\mathcal T_{\mathbb{R}}$, $\mathcal T_{\mathbb{C}}$ be the usual metric topologies on $\mathbb{R}$ and $\mathbb{C}$. Let $\mathbb{R}_{\mathbb{C}}:=\{(a,0): a\in \mathbb{R}\}\subset \mathbb{C}$ and let $\mathcal T_{\mathbb{R}_{\mathbb{C}}}$ be the topology on $\mathbb{R}_{\mathbb{C}}$ obtained from the induced metric. Since the topology induced by the metric topology is the topology obtained from the induced metric we have $\mathcal T_{\mathbb{C}}\mid_{\mathbb{R}_{\mathbb{C}}}=\mathcal T_{\mathbb{R}_{\mathbb{C}}}$. Because $\mathbb{R}_{\mathbb{C}}$ is closed in $\mathbb{C}$, we get $\mathcal T_{\mathbb{R}_{\mathbb{C}}}\subset \mathcal{B}(\mathbb{C})$, and hence $\mathcal{B}(\mathbb{R}_{\mathbb{C}})\subset \mathcal{B}(\mathbb{C})$.
Now, the map $f:(\mathbb{R},\mathcal T_{\mathbb{R}})\to (\mathbb{R}_{\mathbb{C}},\mathcal T_{\mathbb{R}_{\mathbb{C}}})$ defined by $f(a)=(a,0)$ is a homeomorphism. Therefore $f(\mathcal{B}(\mathbb{R}))=\mathcal{B}(\mathbb{R}_{\mathbb{C}})$. Hence we get
$$f(\mathcal{B}(\mathbb{R}))\subset \mathcal{B}(\mathbb{C}),$$
and if we identify $f(\mathcal{B}(\mathbb{R}))$ with $\mathcal{B}(\mathbb{R})$ the result follows.