Choosability in multipartite graph The following is problem from Chapter 5 of Diestel's Graph Theory book:

For positive integers $r,s$, we denote by $K^r_s$ the graph with vertex set
  the disjoint union of sets $V_1,\ldots ,V_r$ of size $s$ and, for all $i\le i < j\le r$, all edges between $V_i$ and $V_j$. Prove that the choice number $\text{ch}(K^n_2) =n$.

Here is what I can prove so far. I try to use induction on $n$. The cases $n=1,2$ are easy to handle. Assume the statement for $n$ and try to prove it for $n+1$. We can prove that $K^{n+1}_2$ is not $n-$choosable by assigning to each of the vertices the list $\{1,\ldots,n\}$ and using the Pigeon-hole principle. Thus it suffices to show $\text{ch}(K^{n+1}_2)\le n+1$.
Suppose we assigned any list of $n+1$ colors to every vertex. It is clear that if some pair of vertices in one of the partite sets contain a common color in their lists then one can exclude that color from the lists of the other $2n$ vertices and apply induction, so we can assume otherwise.
By the same kind of reasoning one can also see that for two vertices which in a partite set if we select a color from one's list and select a color from the other's list, then there is another vertex whose list has both of these colors. Is there anything else I can deduce about the lists?
 A: To prove that $\operatorname{ch}(K_2^n) \le n$, the first step is to do exactly as you do, and reduce to the case where no two vertices in the same part $V_i$ share a color.
Having done so, we can use Hall's theorem to color the graph. Consider the bipartite graph $H$ which, on one side, has the $2n$ vertices of $K_2^n$, and on the other, all the colors appearing on their lists. Join a vertex by edges to all the colors on its list. A perfect matching in $H$ assigns a unique color to every vertex; in particular, it is a proper coloring of the original graph.
To check Hall's condition, consider an arbitrary set $S$ of vertices of $K_2^n$, and compare $|S|$ to $|N_H(S)|$ (the number of neighbors $S$ has in $H$). Whenever $S \ne \varnothing$, we have $|N_H(S)| \ge n$, because each vertex has at least $n$ colors on its list: $n$ neighbors in $H$. So this shows that $|S| \le |N_H(S)|$ whenever $|S| \le n$.
When $|S| > n$, by the pigeonhole principle $S$ must include two vertices from the same pair $V_i$. In that case, those two vertices have no colors in common, so there are at least $2n$ colors on the union on of their lists: $|N_H(S)| \ge 2n$. Since $|S| \le 2n$ because there are only $2n$ vertices, this shows that $|S| \le |N_H(S)|$ in this case as well.
Now, by Hall's theorem, we have a perfect matching, which tells us how to color $K_2^n$, and we are done.
A: $ch(K_2^n)\geq n$.
Proof: Assing to each vertex the same list with colors $\{1,2,\dots,n-1\}$. Clearly no proper list coloring exists.
$K_2^n$ is $n$ choosable.
Proof: If two vertices in the same $V_i$ have the same available color then use this color on them and color the rest inductively.
Otherwise the vertices in the same $V_i$ have the same color, then we can easily prove Hall's condition to color all of the vertices in different color as in Misha's answer.
