Consequence of $V=L$ 
Assume $V=L$. Define $\langle A_\alpha\mid\alpha<\omega_1\rangle$ as follows: Let $A_\alpha$ be the $<_L$-least $A\subseteq\alpha$ such that $(\forall\beta<\alpha)A\cap\beta\not=A_\beta$ if such an $A$ exists and let $A_\alpha=\emptyset$ otherwise. Prove that for all $A\subseteq\omega_1$ there exists an $\alpha<\omega_1$ such that $A\cap\alpha=A_\alpha$.

I've been working on this problem for some time now, and am somewhat stuck. My first thought was to make use of $\diamondsuit$ (since $V=L\Rightarrow\diamondsuit$), but I'm not sure how to apply the principle in this case. Am I wrong to try to apply $\diamondsuit$ here?
 A: As stated, 

Assume $V=L$. Define $\langle A_\alpha\mid\alpha<\omega_1\rangle$ as follows: Let $A_\alpha$ be the $<_L$-least $A\subseteq\alpha$ such that $(\forall\beta<\alpha)A\cap\beta\not=A_\beta$ if such an $A$ exists and let $A_\alpha=\emptyset$ otherwise. Prove that for all $A\subseteq\omega_1$ there exists an $\alpha<\omega_1$ such that $A\cap\alpha=A_\alpha$.

the problem is much less profound than it appears, and doesn't require $\diamondsuit$ or even $V = L$, because:
Claim: $A_\alpha = \emptyset$, for all $\alpha < \omega_1$.
First, note that $\emptyset$ is the $<_L$-first set, period, and $(\forall \beta < 0)\,[...]$ is vacuously true; so $A_0 = \emptyset$, by the first case of the definition ("such an $A$ exists"). [As Noah Schweber points out in his comment, in fact we're done already — there's your "$\alpha$" for the conclusion: $\alpha = 0$.] Now suppose $\alpha > 0$. Then for any $A\subseteq \alpha$, we can't have $(\forall \beta < \alpha)\,  [A \cap \beta \ne A_\beta]$, as $[... \beta ...]$ would have to be true for $\beta = 0$, which it isn't: $A \cap 0 = A\cap\emptyset \,(= \emptyset) \ne A_0 = \emptyset$. So $A_\alpha = \emptyset$, by the second case of the definition ("otherwise").$\quad\square$
Now, if $A\subseteq \omega_1$, $0$ is an $\alpha$ such that $A \cap \alpha = A_\alpha$.
Where is this problem from? Are you sure it's stated correctly?
