This answer is a response by to the answer by Mahmud. I want to show that the construction quoted there is underspecified, and that in at least one complete specification of it the answer is not correct for the general definition of computable numbers from the question. The answer is correct for the usual definition of computable real numbers.
Let $C$ be the set of all Turing machines that are total: for every $T \in C$ and every $n$, $T$ halts on input $n$. It is a standard fact that $C$ is $\Pi^0_2$ complete, so there is no computable enumeration of $C$. Hence the construction in the other answer must choose some noncomputable enumeration of $C$. We construct an enumeration of $C$ so that the real number constructed in the other answer is computable in the limit ($\Delta^0_2$) if our enumeration is used for $C$. Therefore, at best the answer there is underspecified.
(The quoted material is correct in the claim the original author made, which is just that the number constructed is not computable ($\Delta^0_1$); the underspecification only matters when we ask whether the number is computable in the limit ($\Delta^0_2$).)
Motivation: The enumeration we construct is based on an algorithm that uses the halting problem $\emptyset'$ as an oracle. This algorithm constructs the enumeration in stages by a priority argument. At stage $s$ we have a finite list $T^s_1, T_2, \ldots, T^s_s$ of distinct Turing machines. These may or may not be in $C$. At each step, we will extend the list by one, and we may also replace some of the previous elements of the list by new machines. For each location in the list, the machine in that location will be replaced at most once during the entire construction, and if it is replaced it is replaced by a machine in $C$. If a machine is never replaced, then it is also in $C$. We also ensure that every machine in $C$ is put into the list at some stage and never removed. Therefore, the sequence of lists in the construction converges in the limit to an infinite list $C_1, C_2, C_3, \ldots$ which enumerates $C$. Moreover, the construction will ensure that the function $f(n) = C_n(n)$ is computable from $\emptyset'$, which means that the real number obtained by diagonalizing $f$ is also computable from $\emptyset'$. By standard results, this means the diagonalizing function is computable in the limit.
Construction: We fix an effective enumeration of all Turing machines in the background. At stage $0$, use $\emptyset'$ as an oracle to choose the first machine $T$ in the enumeration for which $T(0)$ is defined and let $T_0^0= T$. At stage $s+1$, we have a list $T^s_0\, \ldots, T^s_s$ of machines which, by induction, all halt on inputs $\{0, \ldots, s\}$. Use $\emptyset'$ to ask whether each of these halts on input $s+1$. For each one that does not, replace it by a new machine not in the list which has the same values on $\{0, \ldots, s\}$ and which returns $0$ for all larger values. We can effectively list infinitely many such machines, so we can effectively pick one not in the list, and we can make sure the elements in the list are distinct. Finally, use $\emptyset'$ as an oracle to find the first machine not in the list which halts on all inputs in $\{0, \ldots, s+1\}$ and append it to the list. This ends stage $s+1$.
Verification: As explained above, the limit of this construction gives an enumeration of some infinite set of Turing machine. We prove the set enumerated is exactly $C$. Any machine in $C$ will halt on every input, so it will eventually be added to the list, because at each stage we add the first remaining machine from the original enumeration that halts on enough inputs, and a machine in $C$ halts on all inputs. Conversely, any machine that remains in the list until the end must halt on every input, or else we would remove it. So the machines that remain in the limit are in $C$.
The other key properties of this construction are that:
For each $n$, $C_n(n) = T^n_n(n)$. This is because, when we replace a machine in location $n$, we replace it with one that returns the same value on input $n$.
The function $f(n) = T^n_n(n)$ is computable from $\emptyset'$. This is because the entire construction is computable from $\emptyset'$ (although the limit of the construction is not). So in order to compute $f(n)$ we just simulate the entire construction up to the point where $T^n_n$ is chosen, and then return $T^n_n(n)$.