Let $R= [a,b] \times [c,d]$ be an arbitrary rectangle.
Define $S_1 =\{f(x,y): (x,y)\in R\} $ and $S_2=\{f(x,y):a\leq x \leq b\}$
Claim: For every $c\leq y \leq d$ we want to show that $\inf(S_1) \leq \inf(S_2)$.
My attempt: Let $y_0 \in [c,d]$ be arbitrary.
I realized that by drawing a picture the set $S_2$ the infimum of it is the fixed $y_0 $ we let from the beginning. However, I am not sure how to formally prove this whether if its by definition or by some theorem. Please help, thank in advance.
(1) You have the inequality backwards. For example if the domain of $f$ is $\Bbb R^2$ and $f(x,y)=y^2$ and $R=[0,2]\times [1,2]$ then $\inf S_1=\inf [1,4]=1>0=\inf [0,\infty)=\inf S_2=0.$
(2) $S_1\subset S_2$ so any lower bound for $S_2$ is also a lower bound for $S_1.$ That is, if $A$ is any lower bound for $S_2,$ then for any $z\in S_1$ we have $z\in S_2,$ so $z\ge A.$
So let $L$ be the greatest lower bound of $S_2$.Then $L$ is $one$ $of$ the lower bounds for $S_1,$ so the $greatest$ of the lower bounds for $S_1$ must be be at least $L$.