Assume for every $m\in\mathbb{N}$ there exists bounded sequence of real numbers $\{x_n^m\}_{n=1}^\infty$. Prove that there exists increasing sequence $\{n_k\}$ such that for every $m$ sequence $\{x_{n_k}^m\}_{k=1}^\infty$ converges.
I know that every bounded sequence has converges subsequence. Than for every $m$ we has $\{n_k^m\}$ sequence. Than I want to take these sequences intersection so $\{n_k\}=\bigcap\limits_{m=1}^\infty \{n_k^m\}$ . But I how can I prove that this intersection is infinity? Probably this is the wrong way of proving.
No, it's possible that $\bigcap\limits_{m=1}^{\infty} \left\{n_k^m\right\} = \varnothing$ if you pick sequences arbitrary, consider for example $x_n^1 = (-1)^n$, $x_n^2 = (-1)^{n+1}$, you could pick $n_k^1 = 2k$, $n_k^2 = 2k+1$.
On the other hand, you can pick sequences one by one such that $\left\{x^j_{n_k^m}\right\}$ converges for each $j \le m$. It could be done as follows: let $n_k^0 = k$, and let $\ell_k = n_k^{m+1}$ be the converging subsequence of $\{y_k\} = \left\{x^{m+1}_{n_k^m}\right\}$ (i.e. such that $\left\{y_{\ell_k}\right\}$ converges). Then let $n_k^{m+1} = n_{\ell_k}^m$.
The last step is diagonalization: consider the sequence $z_k = n^k_k$. Consider $\left\{x^j_{z_k}\right\}$. If $k \ge j$ then $z_k \in \left\{n^j_m \colon m \in \mathbb{N} \right\}$ by the properties of the limit $\left\{x_{z^k}^j\right\}_{k = 1}^{\infty}$ converges iff $\left\{x_{z^k}^j\right\}_{k = j}^{\infty} \subseteq \left\{x_{n_k^j}^j\right\}_{k = j}^{\infty}$ converges. The latter converges by our construction.