Cut property
Let $G = (V,E)$ be an undirected graph with an edge weight $w: E \rightarrow \mathbb{R}$. Suppose $X \subset E$ is part of an MST $T$. Let $S \subset V$ be a cut where no edge of $X$ is in the cut edges $cut(S, \overline{S})$. Then for $e^{\ast} \in cut(S, \overline{S})$ of minimum weight - $X \cup \{e^{\ast}\}$ is part of an MST $T^{\ast}$.
Proof
If $e^{\ast} \in T$ then $X \cup \{e^{\ast}\} \subset T$ and we are done.
Assume $e^{\ast} \not \in T$.
Let $e^{\ast} = (a,b)$. As $T$ is an MST it contains a path $p$ from $a$ to $b$. As $a \in S$ and $b \in \overline{S}$ the path $p$ contains an edge $e' \in cut(S, \overline{S})$.
Let $T^{\ast} = T \cup \{e^{\ast}\} \backslash \{e'\}$.
We have the size of $T^{\ast}$ is
$$\vert T^{\ast} \vert = \vert T \vert + 1 - 1 = \vert T \vert = \vert V \vert - 1$$as $T$ is a tree.
Consider the cycle $p e^{\ast}$ - rewrite this cycle as $e' \overline{p}$. We will use this to show $T^{\ast}$ is connected.
Let $a, b \in V$. As $T$ is connected there must a path $p'$ in $T$ connecting $a$ to $b$.
If $p'$ uses $e'$ replace $e'$ by $\overline{p}$ to form $p^{\ast}$ which only uses edges in $T^{\ast}$. As $p^{\ast}$ connects $a$ to $b$ we have $T^{\ast}$ is connected.
As $T^{\ast}$ is connected on $V$ and has $\vert V \vert - 1$ edges it has to be a spanning tree of $G$.
Note as $e^{\ast}$ is the minimum weight edge in $cut(S,\overline{S})$ we have $w(e^{\ast}) \leq w(e')$ and moreover
$$w(T^{\ast}) = w(T) + w(e^{\ast}) - w(e') \leq w(T).$$This gives that $T^{\ast}$ is of minimum weight as $T$ was a MST.
All together this gives us that $T^{\ast}$ is an MST where $X \cup \{e^{\ast}\} \subset T^{\ast}$, proving the statement.