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(Note: This answer works for most any consistient theory, not just $ZFC$.)

We will define a machine $p$ based on the universal algorithm. $p$ does a search, looking for a string that represents a proof of a statement of the form "not ($p$ halts and outputs $n$)" (note that this requires quining, since it is self-referential), for some numeral $n$, such that the proof is valid in ZFC. When it finds such a proof, it outputs $n$.

Lemma: For any numeral $n$, if ZFC is consistient, ZFC does not prove the statement "not ($p$ halts and outputs $n$)".

Assume to the contrary that for some $n$, ZFC proves "not ($p$ halts and outputs $n$)". Then $p$ will find this proof, output $n$, and halt. Moreover, since $p$ must halt in a finite amount of time, we can construct a step by step proof of this in ZFC. That is, we can prove "$p$ halts and outputs $n$" in ZFC. This is a contradiction in ZFC, but ZFC is consistient by assumption. Therefore ZFC does not prove the statement "not ($p$ halts and outputs $n$)" for any numeral $n$.

Now, formalize $p$ into a program $\rho$. Let $$L_\text{ZFC} = |\rho|+1$$ Suppose that $ZFC$ proves for some $s$ that $K(s) \ge L_\text{ZFC} \gt |\rho|$. This implies that the machine that $\rho$ encodes, $p$, does not output $s$. But by the lemma, ZFC can not prove this! Therefore, there is no $s$ such that ZFC proves $$K(s) \ge L_\text{ZFC}$$

Now all that needs to be done is to actually create $\rho$ in some programming language, but we will leave that as an exercise to the programmer.