$\kappa$ is an infinite cardinal, $ \alpha$ an ordinal, while this is the Von Neumann hierarchy:$ \\ \begin{cases} V_0=0 \\ V_{\alpha+1}=P(V_\alpha) \\ V_\lambda=\underset{\gamma<\lambda}{\bigcup}V_\gamma \end{cases}$
$(1\Rightarrow 2) \quad \operatorname{cf}(\alpha)>\kappa$ means trivially that $\alpha$ is a limit ordinal; now, I rewrite $X$ as a set indexed on $\kappa'\leq\kappa, \; X=\{x_i\}_{i\in \kappa'}.$$ \; X\notin V_\alpha \Rightarrow\forall\beta<\alpha \; X\nsubseteq V_\beta$, therefore $\forall\beta\in\alpha \;\; \exists i\in\kappa'\;\;x_i\notin V_\beta\text{ and exists a function } f\colon\underset{i\;\mapsto\;\beta}{\kappa'\to\alpha}\;\;\beta=\min\{\gamma\in\alpha\mid x_i\notin V_\gamma\}$ that is unlimited, but $\operatorname{cf}(\alpha)>\kappa\geq\kappa',\;\text{ so it's absurd.}$
$(\Leftarrow)$ ?
Assume to the contrary that $cf(\alpha)\leq\kappa$. Then there is a cofinal map $c:\kappa\rightarrow\alpha$.
Use this map: for every $\gamma<\kappa$ let $x_\gamma$ be some element of $V_{c(\gamma)}$, and let $X=\{x_\gamma| \gamma<\kappa\}$.
It is not hard to see that $X\subseteq V_\alpha ,|X|=\kappa,$ and $X\notin V_\alpha$ (since $X\notin V_\delta$ for any $\delta<\alpha$).
You should probably also claim that $\alpha$ is a limit ordinal, since if $\alpha=\beta+1$, then $X=\{V_\beta \}$ will be a contradiciton.