Every triangulation decomposes into 3 edge-disjoint spanning trees, when the edges of one triangular face are doubled. See here.
However, if you do not double the edges such a coloring is impossible, since a planar graph has at most $3n-6$ edges (Euler's formula) and three disjoint spanning trees, need $3(n-1)$ edges.
Yes, it is true in general. Note that even though MSTs are non-unique, they are not that non-unique: the only non-uniqueness comes from ties between edges of equal weight. If all edge weights are unique, the MST is also unique.
In particular, from your proof, it follows that if all edge weights are unique, the MST is also minimally connecting.
But suppose you have a general weighted graph $G'$, with possibly non-unique edge weights, and $T$ is one of its MSTs. We can break ties on the edges by adding some very small $\epsilon_i$ to the weight of the $i^{\text{th}}$ edge (taking every $\epsilon_i$ to be much smaller than any actual difference between the cost of any two spanning trees), creating a new graph $G'$ with a unique MST. We can make sure that ties are broken in favor of the edges in $T$ (giving them smaller $\epsilon_i$'s than any edges not in $T$), so that $T$ is also the unique MST of $G'$.
Since $T$ is the unique MST of $G'$, it will be the MST found by Kruskal's algorithm in $G'$, and therefore it is minimally connecting for $G'$ by your definition.
But if we now wipe away all the $\epsilon_i$'s, $T$ should also be minimally connecting for $G$, because $\epsilon_i$'s were much less significant than any actual difference between edge weights. (If some path in $T$ had used an edge of too-high cost in $G$, that edge would have been equally problematic in $G'$.)
Since $T$ was an arbitrary MST, we conclude that all MSTs are minimally connecting.
We can also observe than in a graph with unique edge weights, looking for a minimally connecting tree is forced to find the tree found by Kruskal's algorithm, because we only use an edge of weight $w$ once we've connected all we can by edges of weight $<w$. So in $G'$, MSTs and minimally connecting trees are equivalent, and therefore the same thing is true in $G$. (Start with a minimally connecting tree $T$ in $G$, break ties so that it's minimally connecting in $G'$, conclude that it's an MST in $G'$, and therefore that it's an MST in $G$.)
Best Answer
An edge $e$ crossing $U$ and $V-U$ means that one endpoint of $e$ is in $U$ and another is in $V-U$.
For your statement, we can first let $T$ be a minimum spanning tree containing $X$. As stated, let $e$ be an edge with the smallest weight among those that cross $U$ and $V-U$.
If $e\in T$, then the statement holds trivially.
If $e\notin T$, then consider $T+e$, which contains exactly one cycle $C$, and $C$ crosses $U$ more than once. Therefore, we can have $e'\neq e$ in $C$ with $e'$ crossing $U$. Note that $T'=T-e'+e$ is also a spanning tree, as $T'$ is spanning and connected. Also, $T'$ is a minimum weight spanning tree, since weight of $e$ is not greater than that of $e'$. Hence the statement holds.