Assume that
$\mathcal{B}$ is the set of bounded functions from $\mathbb{R}^d$ to $\mathbb{R}$;
$\mathcal{B}_b$ is the set of bounded Borel measurable functions from $\mathbb{R}^d$ to $\mathbb{R}$;
$C_b$ is the set of bounded continuous functions from $\mathbb{R}^d$ to $\mathbb{R}$.
$C_0$ is the set of continuous functions vanishing at infinity.
Using the knowledge that $\mathcal{B}$ is a Banach space w.r.t. the supremeum norm $\lVert \cdot \rVert_{ \infty }$, I would like to show that the other spaces above are also Banach spaces with the same norm and that
$$ C_0 \subset C_b \subset\mathcal{B}_b \subset \mathcal{B}. $$
I would kindly ask the community members to check whether the solution is correct. I would be grateful if someone points out details that I may have overlooked. I would also appreciate any suggestions that may improve the solution.
So this is how I have proceeded.
First of all, it is obvious that all of the above-mentioned sets are actually vector spaces. Hence consecutively showing that one is a closed subspace of the other should suffice for the proof.
1.
It is obvious that $\mathcal{B}_b \subset \mathcal{B}$. Hence showing that $\mathcal{B}_b$ is a closed supspace of $\mathcal{B}$ is sufficient for concluding that $\mathcal{B}_b$ is a Banach space. Let $(f_n)_{ n \in \mathbb{N}}$ be a sequence in $\mathcal{B}_b$ such that $f_n$ converges in norm ${ \lVert \cdot \rVert }_{\infty}$, i.e. uniformly, to some $f \in \mathcal{B}$. But uniform convergence implies pointwise convergence. Since $f$ is also the pointwise limit of the Borel measurable functinons $f_n$, it is also Borel measurable, and so $\mathcal{B}_b$ is a closed subspace of $\mathcal{B}$ and hence a Banach space.
2.
Since continuous functions from $\mathbb{R}^d$ to $\mathbb{R}$ are Borel measurable, it follows that $C_b \subset \mathcal{B}_b$. To show that $C_b$ is a closed subspace of $\mathcal{B}_b$, again, let $(f_n)_{ n \in \mathbb{N}}$ be a sequence in $C_b$ such that $f_n$ converges in norm ${ \lVert \cdot \rVert }_{\infty}$ to some $f \in \mathcal{B}_b$. To show that $f$ is continuous at any $x_0 \in \mathbb{R}^d$, let $\varepsilon > 0$ be arbitrary. By the continuity of $f_n$, $n \in \mathbb{N}$, we can find some $\delta > 0$ such that for all $| x - x_0 | < \delta$ we have $$ | f_n(x) - f_n(x_0) | < \varepsilon / 3. $$ Moreover, since $f_n$ converges to $f$ uniformly, there is some $n_0 \in \mathbb{N}$ such that for all $n \geq n_0$ we have $$ | f_n(x) - f(x) | < \varepsilon / 3 \quad \text{for all } x \in \mathbb{R}^d \ (\text{including} \ x_0). $$ Therefore, for any $|x - x_0| < \delta$ and $n \geq n_0$ we have \begin{align} | f(x_0) - f(x) | &= | f(x_0) - f(x) + ( f_n(x) - f_n (x_0) ) - ( f_n(x) - f_n(x_0) ) | \\ &\leq | f(x_0) - f_n(x_0) | + | f(x) - f_n(x) | + | f_n(x) - f_n(x_0) | \leq 3 \frac{\varepsilon}{3} = \varepsilon, \end{align} which shows that $f$ is continuous, and so $C_b$ is a closed supspace of $\mathcal{B}_b$ and thus a Banach space.
3.
If $f \in C_0$, then $f$ is continuous and $\lim_{ |x| \rightarrow \infty } f(x) = 0$. To show that $f$ is bounded, fix some $\varepsilon > 0$. Then there is some $M \geq 0$ such that $|x| > M$ implies $|f(x)| < \varepsilon$. Since the closed ball $B_{ M } ( 0 )$ with radius $M$ centered at $0$ is compact, then $f$ is also bounded on it. Therefore $f$ is bounded on the whole $\mathbb{R}^d$. This implies that $C_0 \subset C_b$. To show that $C_0$ is a closed subspace of $C_b$, let $(f_n)_{ n \in \mathbb{N}}$ be a sequence in $C_0$ such that $f_n$ converges in norm ${ \lVert \cdot \rVert }_{\infty}$ to some $f \in C_b$. As shown in the point above, the uniform convergence of the continuous functions $f_n$ to $f$ implies that $f$ is continuous. To show that $\lim_{ |x| \rightarrow \infty } f(x) = 0$, let $\varepsilon > 0$ be arbitrary. Due to the uniform convergence there is some $n_0 \in \mathbb{N}$ such that for all $n \geq n_0$ we have $$ |f_n (x) - f(x)| < \varepsilon/2 \quad \text{for all} \ x \in \mathbb{R}^d. $$ Moreover, for $n = n_0$ there is some $M \geq 0$ such that $|x| > M$ implies $|f_{n_0}(x)| \leq \varepsilon/2$. Thereofore, for $|x| > M$ we have $$ |f(x)| \leq |f(x) - f_{n_0}(x)| + |f_{n_0}(x)| < \varepsilon $$