Local nonsatiation

Question

Suppose that $x^*$ satisfies $x^*\succsim x$ for $\forall x\in\{{x∈X|p·x\leq m}\}$.

How can we prove that $x\succsim x^*$ $\Rightarrow$ $p·x≥m$ if $\succsim$ is locally nonsatiated?

My idea for this goes like this but I am not sure:

Suppose that there exists $x' ∈ X$ such that $x'\succsim x^∗$ and $p · x' < m.$ Since we have local nonsatiation, for any $\epsilon>0$ there exists $x''\in X$ such that $||x''−x'||≤\epsilon$ and $x''\succ x'$. By transitivity of $\succsim$ and by $x''\succ x'\succsim x^*$. The former implies, by $p·x' < m$, that we can choose $\epsilon$ small enough such that $p·x'' < m$ and hence $x''\in\{{x∈X|p·x\leq m}\}$. This implies, by the definition of $x^∗$, that $x^∗\succsim x''$, a contradiction.

Answer 1

By local nonsatiation there exists a sequence $x_n\rightarrow x$ with $x_n\succ x\,\,\forall n$. By transitivity $x_n\succ x^*\,\,\forall n$. By contraposition $p\cdot x_n>m\,\,\forall n$. By continuity of the dot product $p\cdot x\ge m$.

This is pretty much a somewhat more compact version of your own approach.