Problem: Consider the probability space of choosing one point uniformly from $[0,1]$. For $0\leq k\leq n-1$ let $Y_{n,k}$ be the indicator of the interval $\left[\frac{k}{n},\frac{k+1}{n}\right]$. For example, $Y_{5,2}$ is $1$ on $\left[\frac{2}{5},\frac{3}{5}\right]$ and $0$ otherwise. Show that the sequence of random variables $$Y_{1,0},Y_{2,0},Y_{2,1},Y_{3,0},Y_{3,1},Y_{3,2},\dots,Y_{n,0},\dots,Y_{n,n-1},Y_{n+1,0},\dots,$$ converges in probability to $0$, but the sequence does not converge almost surely.
My Attempt: The $j$th term of the sequence will have indices $n,k$ such that $j=2^{n-1}+k$. Note that $P(\vert Y_{n,k}\vert\geq\varepsilon)=\frac{1}{n}$ holds for any $0<\varepsilon\leq1$ and all $n\in\mathbb N$ and $0\leq k\leq n-1$. Therefore, $P(\vert Y_j\vert\geq\varepsilon)=\frac{1}{n}$ where $j=2^{n-1}+k$ as above. It follows that
$$\lim\limits_{j\to\infty}P(\vert Y_j\vert\geq\varepsilon)=0.$$
Since $0<\varepsilon\leq1$ was arbitrary it follows that the sequence converges to zero in probability.
To prove that the sequence does not converge almost surely, observe that for any $\omega\in[0,1]$ there exist infinitely many $n\in\mathbb N$ such that $Y_{n,k}(\omega)=0$ and there also exist infinitely many $n\in\mathbb N$ such that $Y_{n,k}(\omega)=1$. Therefore, the sequence does not converge at $\omega$, and since $\omega\in[0,1]$ was arbitrary, it follows that the sequence does not converge almost surely.
The part of my argument that I am not confident about is the proof of the convergence in probability due to the problem of the indices, since $n$ is not the actual index of the sequence. Does anyone have a suggestion as to how to get around this problem to make the reasoning airtight?
Thank you for your time.