Let $\alpha \in X$. I want to prove following statement $$ \text{Condition } A(\alpha) \text{ is satisfied} \iff \text{Condition } B \text{ is satisfied}. $$
First I assume that $\alpha \in X$ and Condition $B$ is satisfied then I show that Condition $A(\alpha)$ satisfied.
For other direction I assume that $\alpha \in X$ and $A(\alpha)$ is satisfied. Since $A(\alpha)$ is satisfied for arbitrary $\alpha$. It is also satisfied by a concrete $\alpha$. I make some deductions for this concrete alpha and finally reach the conclusion that condition B is satisfied.
Now, have I proven the above proposition correctly?
Your reasoning is correct.
Generally speaking, for any two well-formed formulas $P,Q$, the biconditional formula $P \leftrightarrow Q$ is logicallally equivantent to the formula $(P \to Q) \wedge (Q \to P)$, and if you can show both $P \to Q$ and $Q \to P$ on two separate lines of proof, then you can infer $(P \to Q) \wedge (Q \to P)$ by conjunction introduction.
Given $\alpha \in X$... by first assuming $B$ and then deducing $A(\alpha)$, you derive the conditional formula $B \to A(\alpha)$ by conditional proof (also known as arrow introduction). Likewise, by deducing $B$ under the assumption of $A(\alpha)$, you derive the conditional formula $A(\alpha) \to B$. After having derived both $A(\alpha) \to B$ and $B \to A(\alpha)$ on two separate lines of proof, you may infer $[ A(\alpha) \to B) \wedge (B \to A(\alpha)]$ by conjunction introduciton and conclude $A(\alpha) \leftrightarrow B$ by definition of biconditional.