Define $\mathbb S(n,k)$ as the set of sequences of zeroes and ones of length $n$ with $k$ ones. A crucial fact is that one can draw randomly uniformly an element of $\mathbb S(n,k)$ by conditioning the uniform distribution on the set $\{0,1\}^n$.
To be more precise, consider some i.i.d. random variables $(X_i)_{i\geqslant i}$ uniformly distributed on $\{0,1\}$ and $S_n=X_1+\cdots+X_n$. Then, conditionally on $[S_n=k]$, the random element $(X_i)_{1\leqslant i\leqslant n}$ is uniformly distributed on $\mathbb S(n,k)$.
By symmetry, $x=\mathbb E(X_i\mid S_n=k)$ and $y=\mathbb E(X_iX_j\mid S_n=k)$ do not depend on $i\ne j$ between $1$ and $n$. Summing these and using the fact that $X_i^2=X_i$ for every $i$, one gets $x=k/n$ and $ k^2=n\mathbb E(X_i^2\mid S_n=k)+n(n-1)\mathbb E(X_iX_j\mid S_n=k)=k+n(n-1)y, $ hence $ y=\frac{k(k-1)}{n(n-1)}. $ For every $i\leqslant n$, one gets $\mathbb E(S_i\mid S_n=k)=ix$ and $ \mathrm{var}(S_i\mid S_n=k)=ix+i(i-1)y-i^2x^2=z, $ where $ z=\frac{ik(n-i)(n-k)}{n^2(n-1)}. $ Let $\varepsilon$ denote a positive parameter, and $\mathbb S(n,k,i,\varepsilon)$ the portion of the set $\mathbb S(n,k)$ made of sequences of zeroes and ones of length $n$ with $k$ ones such that the initial fragment of length $i$ has less than $(1-\varepsilon)ik/n$ or more than $(1+\varepsilon)ik/n$ ones.
Then the size of $\mathbb S(n,k,i,\varepsilon)$ divided by the size of $\mathbb S(n,k)$ is exactly the probability of the event $[|S_i-\mathbb E(S_i)|\geqslant\varepsilon ix]$, which, by Bienaymé-Chebychev inequality, is $ \mathbb P(|S_i-\mathbb E(S_i)|\geqslant\varepsilon ix\mid S_n=k)\leqslant\frac{z}{\varepsilon^2i^2x^2}=\frac{A(n,k,i)}{\varepsilon^2}, $ with $ A(n,k,i)=\frac{(n-i)(n-k)}{ik(n-1)}. $ In other words, for every $(n,k,i,\varepsilon)$, the proportion of sequences in $\mathbb S(n,k)$ whose initial portion of length $i$ has a number of ones which differs of its mean value $ik/n$ by more than $\varepsilon ik/n$, is at most $A(n,k,i)/\varepsilon^2$.
Now the asymptotics come into play. Assume that $n\to+\infty$, that $k/n\to p$ for some $p$ in $(0,1)$ and that $i/n\to1/2$. One sees that $nA(n,k,i)\to(1-p)/p$ hence $A(n,k,i)\to0$. This proves the desired result, for every positive $\varepsilon$.