Let $k$ be a field with a norm $|\cdot|$ and $\hat{k}$ its completion regarding this norm (with induced norm). We denote by $|k| \subseteq \mathbb R$ the values of the norm in $\mathbb R$. The question is:
Is the (analytic) closure $\overline{|k|}$ of $|k|$ in $\mathbb R$ equal to $|\hat k|$?
If no, does this hold for non-archimedian norms, i.e. if the norm comes from a valuation?
It is clear that $|\hat k| \subseteq \overline{|k|}$ and equality holds for discrete norms, the question is now given a sequence in $k$ whose norm converges, can we construct a Cauchy-sequence in $k$ with the same norm-limit.
An idea of a counter-example would be $k=\mathbb R(X_1, X_2, \dots)$ the quotient-field over the polynomial ring with countably many variables and valuation $$\nu(X_i)=\lambda_i \in \mathbb Q \text{ with } 1<\lambda_1< \lambda_2 < \dots \text{ and } \lim \lambda_i =\sqrt{42}$$ but I don't see how to prohibit a Cauchy-sequence with the same norm or to construct such one (both should be sufficient for either a proof or refute the statement for valuations). If you prefer you can also choose $(\lambda_i)$ to be a decreasing sequence.
For $k=\overline{\mathbb{Q}}_p$ (the algebraic closure of the field of $p$-adic numbers), $|k|=|\hat{k}|=p^{\mathbb{Q}}\cup\{0\}$. If $\overline{|k|}$ means usual "topologic" closure of $|k|$ (in $\mathbb{R}$), then $\overline{|k|}=\mathbb{R}_{\geqslant 0}$. Does this answer the question?