I am currently going through Munkres's Topology but trying to prove things with nonstandard methods. The one I am struggling with right now is the Lebesgue Number Lemma, not too sure if my proof is correct.
Lebesgue Number Lemma: If the metric space $( X , d )$ is compact and an open cover of $X$ is given, then there exists a number $δ >0$ such that every subset of $X$ having diameter less than $δ$ is contained in some member of the cover.
My proof: Take an infinitesimal number $\delta$ and a set $A \subset X^*$ with diameter less than $\delta$. By definition of compactness, every $x \in X^*$ has shadow in $X$. Let $x,y \in A$ and let $\bar{x}, \bar{y}$ be their shadows in $X$. Then $d(\bar{x}, \bar{y}) \leq d(\bar{x}, x) + d(x, y) + d(y, \bar{y})$. Everything on the RHS is infinitesimal and therefore the LHS, which is standard and infinitesimal, is $0$, i.e., $\bar{x}=\bar{y}$; in other words, every element in $A$ has the same shadow.
Since $A \subset \mu (\bar{x})$, this means that $A \subset $ at least one of the open sets in the open cover that contains $\bar{x}$. In summary, there exists a number $\delta \in (\mathbb{R}^*)^+$ such that for any subset $A \subset X^*$, $A$ is a contained in some member of the open cover. By transfer, we can say there exists a number $\delta \in \mathbb{R}^+$ such that for any subset $A \subset X$, $A$ is contained in some member of the open cover.
Is this proof correct? Thank for any input.
Your proof is correct (up to very minor notational/write-up details). By the way, you can prove the theorem without ever showing that all elements of $A$ have the same shadow:
Consider any metric space $(X,d)$ and open cover $\mathcal{C}$ of $X$.
By Transfer it's sufficient to find some positive $\delta \in ~^\star\!\mathbb{R}$ so that for any $A \in ~^\star\!\mathcal{P}(X) \subseteq \mathcal{P}(~^\star\!X)$ with diameter less than $\delta$, there is some $S \in ~^\star\!\mathcal{C}$ so that $A \subseteq S$.
Take an infinitesimal $\delta > 0$, and take any $A \subseteq ~^\star\! X$ with diameter less than $\delta$, i.e. for all $x,y \in A$, $~^\star\!d(x,y) < \delta$. If $A = \emptyset$ we are done. Otherwise, take some $a \in A$. By the compactness of $X$, we can find $\overline{a} \in X$ so that $~^\star\! d(\overline{a}, a)$ is infinitesimal.
Now, this $\overline{a}$ belongs to some open set $V \in \mathcal{C}$. Consequently, $~^\star\! V \in ~^\star\!\mathcal{C}$ contains all $x \in ~^\star\!X$ such that $~^\star\!d(\overline{a},x)$ is infinitesimal. But since $A$ has diameter less than $\delta$, we have that the distance $~^\star\!d(\overline{a},x) \leq ~^\star\!d(\overline{a},a) + ~^\star\!d(a,x) < ~^\star\!d(\overline{a},a) + \delta$ is infinitesimal for any $x \in A$. Thus $A \subseteq ~^\star\!V$. So we can just set $S = ~^\star\!V$.