[proofplan] The polar condition is a universal inequality over all points of the original set. Since $K \subset L$, every inequality required for membership in $K^\circ$ is already among the inequalities required for membership in $L^\circ$. We prove the inclusion by taking an arbitrary element of $L^\circ$ and checking the defining condition for membership in $K^\circ$. [/proofplan] [step:Take an arbitrary point of the larger polar] Let $y \in L^\circ$. By the definition of the polar of $L$, \begin{align*} x \cdot y \leq 1 \end{align*} for every $x \in L$. [/step] [step:Restrict the defining inequalities from $L$ to $K$] Let $x \in K$. Since $K \subset L$, we have $x \in L$. Therefore the inequality obtained from $y \in L^\circ$ applies to this $x$, and hence \begin{align*} x \cdot y \leq 1. \end{align*} Because $x \in K$ was arbitrary, this inequality holds for every $x \in K$. [/step] [step:Conclude membership in the smaller polar] By the definition of the polar of $K$, the condition \begin{align*} x \cdot y \leq 1 \quad \text{for every } x \in K \end{align*} means exactly that $y \in K^\circ$. Since every $y \in L^\circ$ belongs to $K^\circ$, we have \begin{align*} L^\circ \subset K^\circ. \end{align*} [/step]