Why does $f(x_n)\rightarrow 0$ imply that a subsequence converges to zero of $f$?

Question

Let $f$ be a separable and irreducible polynomial of degree $d\geq 1$ with coefficients in a local field of characteristic $p$, say $K= \mathbb F_p((T))$ and $f\in K[X]$. Assume there is a sequence $(y_n)_{n\in \mathbb N}$ in the separable closure $K^s$ satisfying $$ \lim_{n \rightarrow \infty} f(y_n) = 0. $$ I want to show that there is a subsequence $(y_{n_{k}})_{k\in \mathbb N}$ converging to a zero $x$ of $f$ in $K^s$.
I don't know much about non-archimedian analysis and I wonder if the proof can be stated similar to the real case?

Answer 1

I think one can prove it in the following way:
Let $x_1, \ldots, x_d$ be the different roots of $f$ in $K^s$. Then $f$ can be written in the form $$ f(T)= (T-x_1)\cdots (T-x_d). $$ Assume there is no subsequence $(y_{n_k})_k$ that converges to any zero $x_i$ for $i=1,\ldots ,d$. So we can choose $\varepsilon \gt 0$ and $N\in \mathbb N$ such that for all $n \geq N $ and all $i=1, \ldots , d$ we have $$ \mid y_n - x_i \mid \gt \varepsilon. $$ Therefore $\mid f(y_n) \mid \gt \varepsilon^d \gt 0$ for all $n\geq N$ and we have a contradiction.