Chaitin's incompleteness theorem says no sufficiently strong theory of arithmetic can prove $K(x) > L$ where $K(x)$ is the Kolmogorov complexity of natural number $x$ and $L$ is a sufficiently large constant. The size of $L$ depends on the theory.
Consider the following hierarchy of theories: Let the base theory be Peano arithmetic without induction ($PA^{-}$). Augment $PA^{-}$ with increasingly stronger axioms of simply bounded induction: $I_k : (P(0) \land \forall x_{\leq k}( P(x) \rightarrow P(S(x)) ) ) \rightarrow \forall x_{\leq k}(P(x))$.
Let the Kolmogorov complexity of natural number $x$ be the size of the smallest 2-symbol Busy Beaver (BB) that writes exactly $x$ $1's$ and halts. It is easy to construct a BB with $x$ states that writes $x$ $1's$ and halts. This means $K(x) \leq x$.
(We can also prove $(K(x)<c \rightarrow K(x+1) \leq c)$. Assume we have a BB with $n$ states that writes $x$ $1's$ and halts. We can replace the halt state of this BB with a new state. On $0$ input this state writes $1$ and halts. On input $1$ it writes $1$, moves one position right, and remains in this state. This new BB has one more state than the one for $x$ and writes $x+1$ $1's$ and halts. I don't use this in this argument but it might be useful in some sort of surprise examination argument.)
If PA is consistent it can not prove $K(x) > L(PA)$ where $L(PA)$ is the least natural number such that $PA\not\vdash K(x)\geq L(T)$ for any $x$. Choose a constant much greater than $L(PA)$, $c >> L(PA)$.
Consider the following expression: $\forall x_{\leq k}( K(x) < c )$. We can prove this statement using induction in any theory $PA^- + I_k$ when $k \leq c$. This follows from $K(x) \leq x$. Consider what can be proven in a meta-theory like PA. If PA can not prove $\forall x_{\leq k}( K(x) < c )$ for some $k$ then there is a least such $k$. This means PA proves $K(k) > c$ and is inconsistent. PA must prove $\forall x_{\leq k}( K(x) < c )$ for all $k$. A simple counting argument allows us to prove $\exists k( K(k) > c )$ which is a contradiction. (There is a finite number of BB with size less than $c$).
Where is the flaw in this argument? I know PA must prove $\forall x_{\leq k}( K(x) < c )$ for $k \leq c$ and I am pretty sure it proves it for $k < 2c$.