= Solution
The sampling rule in the PDF is with replacement. Thus conditional on the current proportion $y=j/n$, the next count is $W\sim\operatorname{Bin}(n,y)$. Put $\Delta=W/n-y$ and $q=y(1-y)$. Direct binomial moments give
$$
\mathbb E\Delta=0,\quad\mathbb E\Delta^2=q/n,\quad\mathbb E\Delta^4=3q^2/n^2+q(1-6q)/n^3\leq C/n^2.
$$
The <fourth centered moment of a binomial distribution> implies $\mathbb E|\Delta|^3\leq(\mathbb E\Delta^2\,\mathbb E\Delta^4)^{1/2}\leq Cn^{-3/2}$. The accelerated generator of the <Wright–Fisher binomial sampling chain> is
$$
A_nf(y)=n\mathbb E[f(y+\Delta)-f(y)].
$$
For $f\in C^3([0,1])$, Taylor's formula uniformly yields
$$
\boxed{\sup_{y\in\{0,1/n,\ldots,1\}}|A_nf(y)-\tfrac12y(1-y)f''(y)|\leq C_f n^{-1/2}\longrightarrow0.}
$$
The initial condition is $Z_0^n=\lfloor xn\rfloor$, so $X_0^n\to x$. The endpoints are absorbing.
We give the path-space argument, rather than only a formal generator calculation. Regard $X^n$ as a càdlàg process in $D([0,1],[0,1])$ with the Skorokhod $J_1$ topology. The chain proportion is a bounded <martingale>, with predictable square compensator $V_t^n=n^{-1}\sum_{k<\lfloor nt\rfloor}X_{k/n}^n(1-X_{k/n}^n)$. For bounded <stopping times> $\tau_n\leq\sigma_n\leq\tau_n+\delta_n$, optional sampling gives $\mathbb E|X^n_{\sigma_n}-X^n_{\tau_n}|^2=\mathbb E(V^n_{\sigma_n}-V^n_{\tau_n})\leq(\delta_n+n^{-1})/4$. This bound uses the deterministic number of grid jumps in an interval of length at most $\delta_n$. Compact containment is automatic. The <Aldous tightness criterion> therefore proves tightness.
Furthermore, the fourth-moment bound and a union bound give
$$
\mathbb P\left(\max_{k<n}|X^n_{(k+1)/n}-X^n_{k/n}|>\epsilon\right)\leq C/(n\epsilon^4)\longrightarrow0.
$$
Every subsequential limit is continuous. For each smooth $f$, the discrete compensated process
$$
f(X_t^n)-f(X_0^n)-\frac1n\sum_{k<\lfloor nt\rfloor}A_nf(X_{k/n}^n)
$$
is a <martingale>. Generator convergence and continuity of any limiting path turn the sum into $\int_0^t\tfrac12X_s(1-X_s)f''(X_s)ds$. These compensated processes are uniformly bounded on $[0,1]$, so their <martingale> identities pass to the limit against bounded continuous functions of past coordinates. A <Monotone class theorem> argument extends the identities to the natural filtration. Smooth approximation in the $C^2$ norm extends the test class. Thus every limit solves the <martingale> problem for $Lf(y)=\tfrac12y(1-y)f''(y)$ on $[0,1]$.
To identify this problem with the stated SDE, tests equal to $y$ and $y^2$ on $[0,1]$ (using bounded smooth extensions) give a continuous <martingale> with bracket $\int_0^tX_s(1-X_s)ds$. On an enlargement carrying independent <Brownian motion> $W$, set
$$
B_t=\int_0^t\frac{\mathbf1_{\{X_s(1-X_s)>0\}}}{\sqrt{X_s(1-X_s)}}\,dX_s+\int_0^t\mathbf1_{\{X_s(1-X_s)=0\}}\,dW_s.
$$
Its bracket is $t$, and the <Lévy characterization of Brownian motion> makes it Brownian. The zero-coefficient portion of $dX$ has zero bracket, so $dX_t=\sqrt{X_t(1-X_t)}\,dB_t$. Conversely Itô's formula shows that any such SDE solution solves the limiting <martingale> problem. The allowed uniqueness of the SDE, understood at least as uniqueness in law, therefore makes every subsequential limit have the same law. Consequently
$$
\boxed{X^n\Rightarrow X\text{ in }D([0,1],[0,1]),\qquad dX_t=\sqrt{X_t(1-X_t)}\,dB_t,\quad X_0=x.}
$$
The limit is the <Wright–Fisher diffusion>. Linear interpolation has the same weak limit in the uniform topology because the maximum jump tends to zero. The results applied are Aldous tightness, the vanishing-jump continuous-limit criterion, passage of uniformly bounded <martingale> identities under weak convergence, Lévy characterization, and uniqueness in law; the estimates above verify their hypotheses.
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