Let $X$ be a separable Banach space.
Let $\{x_n\}_n$ be a sequence of $X$ converge to $x$, and for all $n\geq 1$ : $\{y_k^{n}\}_k$ be a sequence converge to $x_n$ when $k$ tends to infinity.
How we can build a sub-sequence of $\{\{y_k^{n}\}_k~:~n\geq 1\}$ which converges to $x$ ? And does the sequence $\{y_n^{n}\}_n$ converge to $x$ when $n$ tends to infinity?
I'm not sure about the first question, but the answer to the second question is negative.
The sequences defined by $\forall n,k \in \mathbb{N}: y_k^n := (\frac{1}{k})^{\frac{1}{n}}$ and $x_n := 0$ in $(\mathbb{R}, \vert \cdot \vert)$ provide a counterexample.
We have $\forall n \in \mathbb{N}: y_n^k \rightarrow 0 = x_n$ for $k \rightarrow \infty$, and $x_n \rightarrow 0$ for $n \rightarrow \infty$. However $y_n^n = (\frac{1}{k})^{\frac{1}{n}} \rightarrow 1 \neq 0$ when $n \rightarrow \infty$.