Equivalent definition of minimal spanning tree
Note: The original answer has a flaw in "without loss of generality". The following one is based on Section 8.2.3 of the book "Computer Algorithms" (3rd edition) by Sara Baase and Allen Van Gelder.



The right part of the theorem is called the MST property.


  MST Property: Let $T$ be any spanning tree. For any edge $e \notin T$, $e$ has the maximal weight on the cycle created by adding it to $T$.


In terms of the MST property, the theorem to prove can be restated as:


  A spanning tree $T$ is an MST $\iff$ $T$ has the MST property.




We first prove the following lemma:


  Lemma: If $T$ and $T'$ are spanning trees that have the MST property, then $w(T) = w(T')$.


Proof: Let $\Delta E = (E(T) \setminus E(T')) \cup (E(T') \setminus E(T))$. That is, $\Delta E$ is the set of edges that are in one of $T$ or $T'$ but not in both. $\Delta E \neq \emptyset$. Let $|\Delta E| = k$.

If $k = 0$, we are done.

Let $k &gt; 0$ and $e$ be a minimum edge in $\Delta E$. Assume that $e \in T \setminus T'$ (the case of $e \in T' \setminus T$ is symmetrical).

$T' + e$ create a cycle $C_{T'}$. There must be an edge $e'$ on the cycle $C_{T}$ that $e' \neq e$ and $e' \in T' \setminus T$ (otherwise $T$ contains the cycle $C_{T'}$).


Since $T'$ has the MST property, $e$ has the maximum weight on the cycle $C_{T'}$. Thus, $w(e') \le w(e)$.
On the other hand, $e$ by definition is a minimum edge in $\Delta E$. Thus, $w(e) \le w(e')$.


Therefore, $w(e) = w(e')$. 

Add $e$ to $T'$, creating a cycle, then remove $e'$, leaving a spanning tree $T''$. Since $w(T'') = w(T')$, $T''$ is an MST. However, $T$ and $T''$ differ by $k-1$ edges.

Repeating the procedure above, we will end with an MST, say $T^{k+1}$, such that $T$ and $T^{k+1}$ differ by $0$ edges.

Therefore, we have $w(T) = w(T^{k+1}) = \cdots = w(T'') = w(T')$.



Given the lemma above, we proceed to prove the "$\Leftarrow$" direction as follows:

Assume $T$ has the MST property. Let $T_m$ be any MST. By the "$\Rightarrow$" direction, $T_m$ has the MST property. By the lemma above, $w(T) = w(T_m)$. Thus, $T$ is an MST.