Quick answer since it is late:
What you want to do is look up ``Baire function'' (in Wikipedia, for example). The Baire class of functions is the least collection of functions $f:\mathbb R\to\mathbb R$ that contains the continuous functions and is closed under limits.
This argument shows that even if you iterate the process (theThe Wikipedia entry talks about class $n$ Baire functions for all $n\in{\mathbb N}$: Baire class one functions are pointwise limits of sequences of continuous functions. Baire class two functions are pointwise limits of sequences of Baire class one functions, etc. You need to go on much longer through a transfinite process), you have to iterate it forcapture all Baire functions, resulting in a hierarchy of length very long time if$\omega_1$. But even then you hopewill fail to capture all functions this way$f:\mathbb R\to\mathbb R$: There are only $\mathfrak c$ many Baire functions. In fact, these are precisely the Borel measurable functions.
Let me addclose with something of an advertisement, now that I have some time. Pete Clark's comments in another answer show that $\chi_{\mathbb Q}$ is not the pointwise limit of continuous functions. For this, he described a characterization of the Baire class 1 functions that clearly $\chi_{\mathbb Q}$ does not satisfy. Since the answer has been deleted, let me repeat Peter's comment:
A real function $f$ is in Baire class one iff for every nonempty perfect subset $P\subset\mathbb R$, there exists $x\in P$ such that the restriction of $f$ to $P$ is continuous at $x$ (in fact, $f\upharpoonright P$ is continuous at all points of $P$ except for a set that is first category in $P$). Taking $P$ to be a closed interval $[a,b]$, this shows that there is a dense subset of points at which $f$ is either left- or right- continuous. This is not the case for $\chi_{\mathbb Q}$.
(Note the nice corollary that, since derivatives are obviously Baire class one functions, they are continuous on a dense set of reals.)
The simplest kind of definability a function mymay have is that its graph is Borel (this is the case if the function is continuous, for example). From here, a very large hierarchy of levels of complexity of subsets of ${\mathbb R}^m$ is defined, starting by taking projections of Borel subsets of ${\mathbb R}^{m+1}$, and complements, and then iterating this procedure.
The fact that we can actually iterate the procedure, i.e., that the hierarchy does not collapse, is where Cantor's diagonalization appears. Anyway, any class of functions with a simple description is easily seen to belong to a (tipicallytypically, very short) initial segment of this hierarchy, and so we know it cannot capture the class of all functions. Many variants of your question are seen immediately to have negative answers through this procedure, which has the advantage of separating levels of complexity in a more refined way than mere cardinality.