Prove that there is a sequence $j_1$x_{i,j}$, $i,j\in \Bbb N$ be real numbers bounded by $1$ i.e., $|x_{i,j}|<1$ for all $i,j \in \Bbb N$. Prove that there is a sequence $j_1<j_2<j_3<\cdots$ such that for every $i$ we have that the sequence $(x_{i,j_n})_{n=1}^\infty$ is convergent.
The way I am thinking it is that for $i=1$ we have $(x_{1,j})$ has a convergent subsequence $(x_{1,j_1})$ then we will consider $(x_{2,j_1})$ which is bounded and hence have a convergent subsequence $(x_{2,j_2})$ then we will consider $(x_{3,j_2})$ and so on inductively we will choose for $i=n$ we get convergent sequence $(x_{n,j_n})$ and then if we take sequences $(x_{1,j_n}),(x_{2,j_n}), \cdots, (x_{n,j_n})$ all of them are convergent but I can't generalise it to for all $i$. Please help here.
 A: Following my comment here is my awnser using diagonlisation. If you've never seen this before it can be a notational nightmare and I'm sorry about that. A tip would be to write out the first few terms of sequences I define below stacked on top of one another and see where the diagonliastion comes from.
We will need Bolzano-Weierstass, which sates any bounded sequence in $\mathbb{R}$ has a convergent subsequence.
As you did in the first step we extract a sequence $(j_n^{(1)})_n$ such that $(x_{1,j_n^{(1)}})_n$ converges. Next notice that $(x_{2,j_n^{(1)}})_n$ is bounded and hence we can again extract a subsequence of $(j_n^{(1)})_n$ call it $(j_n^{(2)})_n$ such that $(x_{2,j_n^{(2)}})_n$ is convergent. But also note that $(x_{1,j_n^{(2)}})_n$ is convergent and converges to the same limit as $(x_{1,j_n^{(1)}})_n$  as it is a subsequence. Repeat this process indefinitely so that we a get an infinite collection of nested sequences. i.e. we get a sequence for each $k\in \mathbb{N}$,  $(j_n^{(k)})_n$ such that $(j_n^{(k+1)})_n \subset (j_n^{(k)})_n$ for all $k$. Then we can extract the "diagonal" sequence (hence the name diagonalisation) which will be precisely the sequence $(j_m^{(m)})_m$ which has tail subsequence of $(j_n^{(k)})_n$ for all $k$. By the construction of our sequences $(j_n^{(k)})_n$ it must follow that $(x_{i,j_m^{(m)}})_m$ must be convergent for all $i$.
Hope this helps!
