This question has a negative answer is many respects.  Firstly, there are simple constructions in the commutative case.  Namely, for $F = K(\!(x)\!)$ we have $F^{\times} = x^{\mathbf{Z}} \times O^{\times}$ where $O = K[\![x]\!]$, so
$${\rm{Hom}}(F^{\times},\mathbf{Z}/p\mathbf{Z}) = (\mathbf{Z}/p\mathbf{Z}) \times
{\rm{Hom}}(O^{\times}, \mathbf{Z}/p\mathbf{Z})$$
(where "Hom" means "continuous homomorphisms), so to show that the commutative profinite $O^{\times}$ is *not* topologically finitely generated it is sufficient to show that the left side above is infinite.  But by local class field theory the left side is ${\rm{H}}^1(G_F,\mathbf{Z}/p\mathbf{Z})$ (continuous cohomology), and by Artin-Schreier this is $F/\wp(F)$ where $\wp(f) = f^p-f$.  Consideration of polar parts in $F = K(\!(x)\!)$ shows that $F/\wp(F)$ is infinite.  Thus, the group $K[\![x]\!]^{\times}$ is not topologically finitely generated.  Likewise, via the surjection $$\det:{\rm{GL}}_n(K[\![x]\!]) \rightarrow K[\![x]\!]^{\times}$$
the same happens for any $n$.

But this is really cheating: clearly the "correct" question should get away from the silly constructions that one can make with ease in the commutative setting, so one ought to replace GL$_n$ with semisimple groups $G$ over $O = K[\![x]\!]$.  It is also a bad idea to consider general closed subgroups, since with Borel subgroups the torus factor allows us to use the same silly commutative constructions; e.g., if $G = {\rm{SL}}_n$ as an $O$-group with $n > 1$ and $B \subset G$ is the standard Borel $O$-subgroup then $B(O)$ has the $(n-1)$-fold direct product of copies of $O^{\times}$ as a quotient, so obviously $B(O)$ is not topologically finitely generated.

Likewise, if $G$ is not *simply connected* then more commutative silliness gets in the way.  For example, the determinant induces ${\rm{PGL}}_n(O) \twoheadrightarrow O^{\times}/(O^{\times})^n$, so for $n$ divisible by $p$ we have the same problem once again (as $O^{\times}/(O^{\times})^p$ is a quotient, and it has infinitely many continuous homomorphisms to $\mathbf{Z}/p\mathbf{Z}$ as we have seen above).

So finally it seems that the version of the question not easily falsified by commutative tricks is for $G$ to be a semisimple $O$-group that is simply connected and to consider the *open* subgroups of $G(O)$ (which are of course automatically closed).  It is a general fact (via deformation theory of semisimple groups) that any such $G$ is the scalar extension along $K \rightarrow O$ of its reduction, though you might wish to just consider only such $G$ without knowing it is the most general case.  By coming from $K$ we see that $G$ arises from a "constant" $K$-group over the global ring $K[x]$.  One can then use strong approximation for *simply connected* groups over global fields (such as $K(x)$) and results of Behr on finite generation of $S$-arithmetic groups when the sum of local ranks at $v \in S$ is at least 3) to deduce that $G(O)$ contains a dense finitely generated subgroup, namely an $S$-arithmetic group for suitable $S$, so $G(O)$ is topologically finitely generated.  Moreover, any open subgroup of $G(O)$ is defined by a "congruence condition", so one can still find $S$-arithmetic groups dense in such open subgroups, and hence they're again topologically finitely generated.