Tropical Lifting Lemma (Puiseux Series) The general result in tropical geometry is
$K$ algebraically closed valued field
$I$ ideal of $K[x_1, \cdots, x_n]$ $V(I) = \lbrace \bar{a}\in K^n: f(\bar{a})=0 \text{ for all } f \in I \rbrace$. 
Let $f \in K[x_1,\cdots,x_n]$ such that $f(X) = \sum a_j X^j$ $j \in \mathbb{N}^n$ then $\operatorname{Trop}(f): \mathbb{R}^n \rightarrow \mathbb{R}$ $X\mapsto \max\lbrace-v(a_j)+\langle X,j\rangle\rbrace$  where $v$ is the valuation map.
A root of $\operatorname{Trop}(f)$ is a tuple $w$ such that the maximum of $\lbrace -v(a_j) +\langle w, j \rangle \rbrace$ is attained at least twice. 
The fundamental theorem of tropical geometry states $v(V(I)) = \operatorname{Roots}(\operatorname{Trop}(I))$ (actually $\text{the topological closure of } v(V(I)) = \operatorname{Roots}(\operatorname{Trop}(I))$, or $v(V(I)) = \operatorname{Roots}(\operatorname{Trop}(I))\cap \mathbb{Q}^n$  in the case of the Puiseux field). Trivially $v(V(I)) \subseteq \operatorname{Roots}( \operatorname{Trop}(I))$, nevertheless the converse is not so easy to prove.
My question goes on how to prove the converse for $K$ the field of Puiseux series. Maybe an approach through Hensel's Lemma. 
 A: Several people (including me) have written down flawed proofs. The awkwardness is that the valuation group is not discrete. This makes things easier, in that there are fewer obstructions to lifting in that setting, but it makes things harder in that you can't cite standard references on valuation rings when you want to. In particular, this is why you can't just say "Hensel" and be done.
I think it is now generally agreed that the first fully correct proof is Sam Payne's. I also see that Sam has a preprint with Brian Osserman which looks relevant, but I haven't read it yet.

Regarding the question about whether the case of $K=\bigcup \mathbb{C}((t^{1/n}))$ is easier than the general case: There are two parts to proving the theorem for this special case. The first is to show that $K$ is algebraically closed. There is a nice exposition of this in Section 1.12 of Kollar's "Lectures on Resolution of 
Singularities". This is, indeed, simpler than the corresponding proofs in the characteristic $p$ case.
The second part is to, more or less, show that we can lift solutions from the special fiber to the general fiber. This is a Hensel's lemma type step, where at a key point you need to use that $K$ is algebraically closed. This is the part which tends to be difficult to do correctly.
What you are asking is whether the fact that it is easier to prove $K$ is algebraically closed, in the case you describe, makes the proof easier as a whole. I don't think it is a big improvement, because the technical part is in using that $K$ is algebraically closed, not proving it. But if you find every part hard, then simplifying this part might give you one less thing to think about.
