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Here is a sketch of how you might approach this. Let $$\xi_j = \frac{\kappa(I_{j-1} - X_{j-1})} {\sqrt{X_{j-1}(1 - X_{j-1})}}.$$ The two "hard" results that must be proven are: (1) For each $t$, $$\lim_{\kappa\to0} E\big[\max\{|\xi_j|: 1 \le j \le \lfloor\kappa^{-2}t\rfloor\}\big] = 0,$$ and (2) for each $t$, $$\sum_{j=1}^{\lfloor\kappa^{-2}t\rfloor} \xi_j^2 \to t,$$ in probability as $\kappa\to0$. The rest of the proof would then be the following "soft" argument based on general theory.

First, let $W^\kappa(t)=\sum_{j=1}^{\lfloor\kappa^{-2}t\rfloor} \xi_j$. Using the two results above, one can use the martingale central limit theorem (Theorem 7.1.4 in Ethier & Kurtz) to prove that $W^\kappa\Rightarrow W$, where $W$ is a standard Brownian motion.

Next, we take the difference equation which defines the sequence ${X_n}$ and rewrite it as an integral equation. More specifically, if we define $X^\kappa(t)=X_{\lfloor\kappa^{-2}t\rfloor}$, then we may write $$X^\kappa(t) = X_0 + \int_0^t \sqrt{X^\kappa(s-)(1 - X^\kappa(s-))}\,dW^\kappa(s).$$ There is nothing deep here, just a change of notation, really.

Finally, we use Theorem 5.4 in Kurtz and & Protter to prove that $(X_0,X^\kappa,W^\kappa) \Rightarrow(X_0,X,W)$, where $X$ is the unique strong solution to $dX=\sqrt{X(1-X)}\,dW$, $X(0)=X_0$.

A watered-down version of Theorem 5.4 in Kurtz and & Protter is available as Theorem 2.3 in these lecture notes. This version is sufficient for your purposes, and it may be easier to digest. Also, to use this theorem, you must show that, for every version of $(X_0,W)$, the limiting SDE has a unique strong solution for all time. This follows, for example, from Proposition 5.2.13, Theorem 5.5.4, and Corollary 5.3.23 in Karatzas and & Shreve.

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Here is a sketch of how you might approach this. Let $$\xi_j = \frac{\kappa(I_{j-1} - X_{j-1})} {\sqrt{X_{j-1}(1 - X_{j-1})}}.$$ The two "hard" results that must be proven are: (1) For each $t$, $$\lim_{\kappa\to0} E\big[\max\{|\xi_j|: 1 \le j \le \lfloor\kappa^{-2}t\rfloor\}\big] = 0,$$ and (2) for each $t$, $$\sum_{j=1}^{\lfloor\kappa^{-2}t\rfloor} \xi_j^2 \to t,$$ in probability as $\kappa\to0$. The rest of the proof would then be the following "soft" argument based on general theory.

First, let $W^\kappa(t)=\sum_{j=1}^{\lfloor\kappa^{-2}t\rfloor} \xi_j$. Using the two results above, one can use the martingale central limit theorem (Theorem 7.1.4 in Ethier & Kurtz) to prove that $W^\kappa\Rightarrow W$, where $W$ is a standard Brownian motion.

Next, we take the difference equation which defines the sequence ${X_n}$ and rewrite it as integral equation. More specifically, if we define $X^\kappa(t)=X_{\lfloor\kappa^{-2}t\rfloor}$, then we may write $$X^\kappa(t) = X_0 + \int_0^t \sqrt{X^\kappa(s-)(1 - X^\kappa(s-))}\,dW^\kappa(s).$$ There is nothing deep here, just a change of notation, really.

Finally, we use Theorem 5.4 in Kurtz and Protter to prove that $(X_0,X^\kappa,W^\kappa) \Rightarrow(X_0,X,W)$, where $X$ is the unique strong solution to $dX=\sqrt{X(1-X)}\,dW$, $X(0)=X_0$.

A watered-down version of Theorem 5.4 in Kurtz and Protter is available as Theorem 2.3 in these lecture notes. This version is sufficient for your purposes, and it may be easier to digest. Also, to use this theorem, you must show that, for every version of $(X_0,W)$, the limiting SDE has a unique strong solution for all time. This follows, for example, from Proposition 5.2.13, Theorem 5.5.4, and Corollary 5.3.23 in Karatzas and Shreve.