[proofplan]
We prove both implications directly from the definition of Turing degree. If $A$ is computable, then an ordinary decision procedure for $A$ can be used as an oracle computation relative to any oracle, so $A \leq_T \varnothing$; the reverse reduction $\varnothing \leq_T A$ is automatic because the empty set is computable without consulting the oracle. Conversely, if $A \leq_T \varnothing$, an oracle machine computing $A$ from the empty oracle can be simulated by an ordinary machine by answering every oracle-membership query negatively, since no natural number belongs to $\varnothing$.
[/proofplan]
[step:Show that a computable set has the same Turing degree as the empty set]
Assume $A \subseteq \mathbb{N}$ is computable. Let
\begin{align*}
\mathbf{1}_A: \mathbb{N} &\to \{0,1\}
\end{align*}
be the characteristic function of $A$, defined by $\mathbf{1}_A(n)=1$ if $n \in A$ and $\mathbf{1}_A(n)=0$ if $n \notin A$. Since $A$ is computable, there is an ordinary Turing machine $M_A$ which halts on every input $n \in \mathbb{N}$ and outputs $\mathbf{1}_A(n)$.
View $M_A$ as an oracle Turing machine which never asks an oracle query. With oracle $\varnothing$, this machine still halts on every input $n$ and outputs $\mathbf{1}_A(n)$. Hence $A \leq_T \varnothing$.
It remains to verify $\varnothing \leq_T A$. Define
\begin{align*}
\mathbf{1}_{\varnothing}: \mathbb{N} &\to \{0,1\}
\end{align*}
by $\mathbf{1}_{\varnothing}(n)=0$ for every $n \in \mathbb{N}$. The ordinary machine which outputs $0$ on every input computes $\mathbf{1}_{\varnothing}$, and therefore also computes it relative to the oracle $A$ without asking any oracle query. Thus $\varnothing \leq_T A$.
Since $A \leq_T \varnothing$ and $\varnothing \leq_T A$, the sets $A$ and $\varnothing$ are Turing equivalent. Therefore $\deg_T(A)=\deg_T(\varnothing)=0$.
[/step]
[step:Simulate an empty-oracle computation by an ordinary computation]
Assume $\deg_T(A)=0$. By the definition of the computable degree, this means $A$ is Turing equivalent to $\varnothing$, so in particular $A \leq_T \varnothing$. Thus there exists an oracle Turing machine $M^{(\cdot)}$ such that, when run with oracle $\varnothing$, it halts on every input $n \in \mathbb{N}$ and outputs $\mathbf{1}_A(n)$.
Define an ordinary Turing machine $N$ as follows. On input $n \in \mathbb{N}$, the machine $N$ simulates the computation of $M^{\varnothing}$ on input $n$. Whenever the simulated oracle machine asks whether a number $m \in \mathbb{N}$ belongs to its oracle, the machine $N$ answers $0$. This answer is correct for the oracle $\varnothing$, because $m \notin \varnothing$ for every $m \in \mathbb{N}$.
Therefore the computation path of $N$ on input $n$ is exactly the computation path of $M^{\varnothing}$ on input $n$. Since $M^{\varnothing}$ halts on every input and outputs $\mathbf{1}_A(n)$, the ordinary machine $N$ also halts on every input $n$ and outputs $\mathbf{1}_A(n)$. Hence $\mathbf{1}_A$ is computable, so $A$ is computable.
[guided]
We must turn an oracle computation relative to $\varnothing$ into a computation with no oracle at all. The hypothesis $\deg_T(A)=0$ says that $A$ is in the same Turing degree as the empty set. In particular,
\begin{align*}
A \leq_T \varnothing.
\end{align*}
By the definition of Turing reducibility, there is an oracle Turing machine $M^{(\cdot)}$ such that $M^{\varnothing}$ halts on every input $n \in \mathbb{N}$ and outputs the value $\mathbf{1}_A(n)$.
The only obstruction to making $M^{\varnothing}$ into an ordinary machine is that $M^{(\cdot)}$ may ask oracle queries. But the oracle is $\varnothing$, and membership in $\varnothing$ has a fixed answer: for every queried number $m \in \mathbb{N}$,
\begin{align*}
m \notin \varnothing.
\end{align*}
So every oracle query receives the answer $0$.
Define an ordinary Turing machine $N$ as follows. Given input $n \in \mathbb{N}$, the machine $N$ simulates $M^{(\cdot)}$ on input $n$. If the simulated machine asks whether $m \in \varnothing$, the machine $N$ writes the answer $0$ and continues the simulation. This produces exactly the same sequence of configurations that $M^{\varnothing}$ would produce, because every oracle answer supplied by $N$ is the correct empty-oracle answer.
Since $M^{\varnothing}$ halts on every input $n$ and outputs $\mathbf{1}_A(n)$, the simulating machine $N$ also halts on every input $n$ and outputs $\mathbf{1}_A(n)$. Thus the characteristic function
\begin{align*}
\mathbf{1}_A: \mathbb{N} &\to \{0,1\}
\end{align*}
is computed by an ordinary Turing machine. Therefore $A$ is computable.
[/guided]
[/step]
[step:Conclude the equivalence]
The first step proves that computability of $A$ implies $\deg_T(A)=0$. The second step proves that $\deg_T(A)=0$ implies computability of $A$. Therefore, for every $A \subseteq \mathbb{N}$,
\begin{align*}
\deg_T(A)=0 \iff A \text{ is computable}.
\end{align*}
[/step]
