There seems to be a lot of confusion in the other answers and comments thereto. I'll try to make things more clear. (In particular, @Holowitz's and @Nico's answers are incorrect, as I showed in the comments below @Nico's answer.)
First note that a real number can be identified with a set of natural numbers by declaring that the binary expansion of the real corresponds to the characteristic sequence set of natural numbers. (Of course, some reals have non-unique binary expansion, so this identification doesn't always make sense, but any such real gives rise to a computable set no matter which binary expansion is taken, so insofar as we're concerned, this is irrelevant.)
Now, the definition of computable you gave is what's known as limit computable or $\Delta_{2}^{0}$, though since the notation $\Delta_{2}^{0}$ is a priori reserved for sets defined by formulas that are both $\Sigma_{2}^{0}$ and $\Pi^{0}_{2}$ in the arithmetical hierarchy, it must be proven that the two notions coincide. This is exactly Schoenfield's limit
lemma together with Post's theorem.
Regarding your request for a "concrete" example of a real that is not $\Delta^{0}_{2}$ (i.e. one that can be "picked by defining it"), the number--let's call it $\alpha$--given in the paper cited in the answer you accepted is defined only relative to some fixed (incomputable!) enumeration of all $\Delta^{0}_{2}$ reals; different enumerations give rise to different $\alpha$'s. However, such an enumeration is inherently not $\Delta^{0}_{2}$ (it can be $\Delta^{0}_{3}$ as @Carl points out in the comments to @Mahmud's answer) in the arithmetical hierarchy, so I do think that it is an acceptable example (contrary to my comments) . Given that, though, it's evident that a much better (i.e. less complex) real can be given: take any strictly $\Sigma^{0}_{2}$ or $\Pi^{0}_{2}$ real. For example, the set of indices of total computable functions $\mathrm{Tot}:=\{e:\forall n \exists s \, [\varphi_{e,s}(n)\downarrow] \}$ is a well-known $\Pi^{0}_{2}$-complete real.
The moral of the story is that there are many arithmetically definable reals that are not "computable" in the sense you defined; just take any arithmetical real that is not $\Delta^{0}_{2}$.
No. For any $n$, define
$a_n =
\begin{cases}
\pi/2^r, & \text{if Turing machine $n$ halts in $r$ steps on input $n$} \\
0, & \text{if Turing machine $n$ never halts on input $n$}
\end{cases}
$
$a_n$ is computable; we can approximate it to any desired accuracy. But if you could determine whether $a_n$ was transcendental or not, you would have solved the Halting Problem.
Best Answer
Using this definition of left-computability, our aim is to construct a computable increasing sequence of rationals that converges to a non-computable number.
Consider a sequence $q_{n}$ subunitary rational numbers.
Define the $k$'th digit after the decimal point of $q_{n}$ to be $1$ if $k \leq n$ and the program encoded by $k$ halts after $n$ steps or less, or $0$ otherwise.
The $q_{n}$ are increasing, as a digit that becomes $1$ stays $1$ for all $q_{p}$ with $p>n$. They are also rational, as each $q_{n}$ has at most $n$ non-zero decimal places. However, their limit will be a number $q$ who's $k$th digit is $1$ if the program encoded by $k$ halts, or $0$ otherwise. As it encodes the solution to the halting problem, it is not computable.
Thus $q$ is left-computable, but not computable.