Let $G$ a simply connected group over $\mathbb{C}$ and $Gr:=G(\mathbb{C}((t)))/G(\mathbb{C}[[t]])$ the affine grassmannian. By Cartan decomposition we have a partition of stratas indexed by $\lambda\in X\_{\*}(T)^{+}$.
Let $Gr\_{\lambda}$ such a strata and $\overline{Gr}\_{\lambda}$ the closure in $Gr$.


Then, $\overline{Gr}\_{\lambda}=\coprod\limits\_{\mu\leq\lambda}Gr\_{\mu}$, is it a Whithney stratification?



Whitney stratification and affine grassmanian
The point is following: $\overline{Gr^\lambda}$ is a finite dimensional variety acted upon by the pro-algebraic group $G(\mathbb{C}[[t]])$.


This action factors through the action of some finite dimensional algebraic group called $G(\mathcal{J}^l)$. The strata you describe are just the orbits of this group action and there are finitely many of them. (See explanation below)


It is well known, that if a complex algebraic group $G$ acts on a complex variety $X$ algebraically with only finitely many orbits, then the $G$-orbits form a Whitney stratification of $X$, see [Algebraic Stratifications of $G$-varieties](https://mathoverflow.net/questions/129218/algebraic-stratifications-of-g-varieties/129806#129806) for details.


Remark: Let $\mathcal{J}^l:= \mathbb{C}[[t]]/t^l \mathbb{C}[[t]]$ for $l\in \mathbb{N}$. Then $G(\mathcal{J}^l)$ is a finite dimensional algebraic group. Let $G(\mathcal{O}^l)$ be the kernel of the canonical homomorphism $G(\mathbb{C}[[t]])\to G(\mathcal{J}^l)$. 
If you choose an embedding $G\hookrightarrow GL\_n(\mathbb{C})$ you can think of $G(\mathcal{O}^l)$ as the subgroup of $G(\mathbb{C}[[t]])$ where all entries on the diagonal are in the form of $1+t^l\cdot f$ with $f\in \mathbb{C}[[t]]$ and all other entries are in the form of $t^l f$. Anyway it is straight forward to show that for any $\lambda$ there exists an $l\gg 0$ such that $G(\mathcal{O}^l)$ operates trivially on $\overline{Gr^\lambda}$, in other words the $G(\mathbb{C}[[t]])$ action factors through
 $G(\mathcal{J}^l)$


