A Question Regarding Diagonalization For this question I am only considering binary sequences of countably infinite length.  
Consider an arbitrary set $S$ of such sequences, $S$ of order type omega, $\omega$.  By diagonalization one can construct a binary sequence $s_*$ not contained in $S$. Add $s_*$ to $S$ to form a new set, call it $S'$, of countable order type $\omega+1$.  Now it would seem that one can diagonalize out of $S'$ and construct another binary sequence, call it $s_*'$ ($s_*'$ of order type $\omega$) contained in neither $S$ nor $S'$ and continue this process until the order type of the set formed by the operations of diagonalization and adding the diagonal (which is always of order type $\omega$) to the set is at least $\omega_1$.  
Can this process be continued past $\omega_1$?  If not, why not?  If yes, then how?          
 A: First of all, when you say "order type" it implies that there is a natural order on the sequences. That doesn't necessary happen. Moreover the resulting sequence from the diagonal argument need not be larger in the order that you chose.
But suppose that you begin with a countable set, then using diagonalization you add more and more elements. Can you go beyond $\omega_1$? Well, not necessarily. If the continuum hypothesis holds, or if we are in a context of $\sf ZF$ and $\aleph_1\nleq2^{\aleph_0}$, then it might be the case that your process has exhausted itself somehow.
Either by showing that this inductive method is insufficient to define a set of size $\aleph_1$ (i.e. we have to make some uncountably many choices (in the context of $\sf ZF$ it might be the case that going through all the $\omega_1$ steps requires a scale to $\omega_1$, that is a sequence of enumerations for all the countable ordinals, which may not exist without the axiom of choice); or that we simply lucked out and managed to cover all the real numbers, in the case that $2^{\aleph_0}=\aleph_1$, in which case you can't continue anymore.
If you could have proved from $\sf ZF(C)$ that diagonalization carries to ordinals of size $\aleph_1$, then you would have proved $\lnot\sf CH$ and so the inconsistency of $\sf ZFC$. However if we assume more, e.g. $\sf MA+\lnot CH$, then we can switch to other techniques to produce newer sequences.
