Theorem:
Any graph $G$ has two sided subgraph $H$, where $|E(H)| \ge {|E(G)| \over2}$
Proof: Taking two any sets $U$ and $W$ , so that $U \cup W = V(G)$. It is clear that any graph has a subgroup ,where vertices of subgraph is equal to vertices of graph. By ejecting edged in $U$ , where any vertex is connected to other, and doing same for $W$, we get subgraph $H$. Now showing that for any $u \in V(G)$, $d_H(v) \ge {d_G(v) \over 2}$. Suppose the opposite is true, $\exists u \in V(G)$, so that $d_H(v) < {d_G(v) \over2}$. Suppose $ u \in U$.
$d_H(v) < {d_G(v) \over2}$ => $u$ vertex has more neighbor vertices in $U$ , than in $W$.
I stuck here, don't have any idea , what is written in last line, any helps will be appreciated.
Why not use induction to prove the claim. Remove a vertex $v$ from $G$. Now let $H_v$ be a bipartite subgraph of $G \setminus \{v\}$ with at least half the edges of $G \setminus \{v\}$ assume by induction there is such a $H_v$. Then writing as $U_v$ and $W_v$ the two parts of $H_v$, we note that for some $a_1$ and $a_2$ that $v$ has $a_1$ neighbors in $U_v$ and $a_2$ neighbors in $W_v$. Assume WLOG that $a_1 \geq a_2$. Then $H = H_v$ plus the $a_1$ edges between $v$ and $U_v$, is bipartite with parts $U_v$ and $W_v + \{v\}$ (indeed, one side $U_v$ of $H$ is independent and the other side $W_v + \{v\}$ of $H$ is also independent as the $a_2$ edges between $v$ and $W_v$ were removed), and has at least half the number of edges of $G$.
[I assume by "two-sided" you mean bipartite? I am not sure how to modify your argument, as for one thing I don't know what you mean by "any graph has a subgroup"]