Let $B\subset A = \{1,2,3,...,99,100\}$ and $|B|= 48$. Prove that exist $x,y\in B$, $x\ne y$ such that $11\mid x+y$. 
Let $B\subset A = \{1,2,3,...,99,100\}$ and $|B|= 48$.
Prove that exist $x,y\in B$, $x\ne y$ such that $11\mid x+y$.

Proof: Let  $P_0:= \{11,22,...,99\}$ and for $i=
1,2,...49$ and $11\nmid i$ make pairs $P_i:= \{i,99-i\}$. Now we
have $46$  subsets with sum of each pair in each subset divisible
by $11$. So if we took at most $1$ element from each set $P_i$ in to $B$
then $B$ would have at most $47$ elements (if $100$ is in $B$ also).
A contradiction.
This bound is sharp. Take $$B:=\{i\in A; i\equiv x\pmod{11},\,x\in \{1,2,3,4,5\}\} \cup \{11\}$$ Then $|B| =47$ and there is no $x,y\in B$ such that $11\mid x+y$.

My question here is:
Is this problem doable by polynomial method? As to watch in $\mathbb{Z}_{11}[x,y]$ a polynomial $$p(x,y) := \prod_{i=1}^{10}(x+y-i)$$ If the statment would not hold, then $p(x,y)=0$ for all $(x,y)\in \{(a,b)\in B' \times B'; \;a\ne b\}$ where $B' = B\pmod {11}$. Clearly $|B'|\geq 6$. Is this good idea?

I'm not sure if I understand your question correctly since it seems to me that you almost got the proof by the combinatorial nullstellensatz (?). Anyway, I will write down my proof here:
First, if there are more than two multiples of $11$ in $B$ then we are done. Now suppose that there is at most one multiple of $11$ in $B$ and consider $C$ is the subset of $B$ removing the multiple of $11$ (if there is) in $B$, so $|C| \ge 47$. By taking modulo $11$, we see that if $C' = C \pmod {11}$ then clearly, $|C'| \ge 6$.
Now consider the polynomial $$p(x,y) := \prod_{i=1}^{10}(x+y-i)$$ for any $x,y \in C'$.
This polynomial has degree $10$ and the coefficient of $x^5y^5$ is ${10 \choose 5} \not = 0 \pmod {11}$; also $|C'| = 6 > 5$ so there is $a,b \in C'$ such that $p(a,b) \not = 0 \pmod{11}$, i.e, $a+b = 0 \pmod{11}$. Note that both $a$ and $b$ are not $0 \pmod{11}$, so $a\not = b \pmod{11}$. Hence, the claim follows.
