I was trying to prove that if $\vert Y \vert \le c$ a.s. then $ \vert E\{Y |G\}\vert \le c$ a.s. also by contradiction.
If I assume $\vert Y \vert \le c$ a.s and $P(\vert E\{Y |G\}\vert > c)>0$ then
$P(\vert E\{Y |G\}\vert > c,c\ge \vert Y \vert)>0$ since $\vert Y \vert \le c$ a.s. and we can conclude that $P(\vert E\{Y |G\}\vert >\vert Y \vert)>0$.
Another alternate idea that I had was Since conditional expectation E{Y |G} takes the values $E[Y\mathbb{1}_A]$ as $A$ varies in $G$ we can conclude that $Y\mathbb{1}_A \le Y \le \vert Y \vert \le c $ we have that for all $A$ in $G$ we have $E[Y\mathbb{1}_A] \le c$ and hence $\vert E\{Y |G\} \vert \le E\{\vert Y \vert |G\} \le c$
Your argumentgs don't seem rigorous. For a correct proof you have to use the definition of conditional expectation. Let $Z=E[Y|\mathcal G]$. By definition $\int_AZdP=\int_A YdP$ for all $A \in \mathcal G$. Let $A=\{Z >c\}$. If $P(A) >0$ then $c P(A) <\int_A Z dP =\int_A YdP \leq cP(A)$ which is a contradiction. This proves that $P(A)=0$ so $Z \leq c $ a.s.. Similarly $Z >-c$ a.s.