BST - Sufficient condition for connecting a node to a parent
Due to the lack of definitions in the question, let us pick up a couple of definitions that are needed for a formal proof.


  A binary rooted tree is a BST if its nodes are visited in its natural order by the inorder traversal. Inserting a new node into a BST means adding a new edge so that the new node becomes the child of some old node and the tree remains a BST.




Let us prove a useful lemma.
(Uniqueness of insertion into BST) Let $T$ be a BST. Let node $X$ be a new node that is not equal to any node in $T$. Then there exists a unique way to insert $X$ into $T$ so that $T$ remains a BST with the same root, where the insertion must add one edge to $T$ (without modifying the existing edges).

Proof by mathematical induction on $n$, the number of vertices in $T$.


The base case when $n=1$. If the new node is less than the root node, it has to inserted as the left child of the root node; otherwise, it has to be inserted as the right child of the root node.
Assume the lemma is true for BST whose number of vertices is smaller than $n$. Suppose $T$ is a BST of $n$ vertices with node $R$ as its root node. Suppose $X$ is a new node that is not equal to any node of $T$. There are two cases.


$X$ is less than $R$. If $R$ has no left child, then $X$ has to be inserted as the left child of $R$. Otherwise, $R$'s left subtree is a BST (because of the recursive nature of the inorder traversal). $T$ has to be inserted into $R$'s left subtree. By induction hypothesis, there exists a unique way to insert $X$ into $R$'s left subtree.
$X$ is greater than $R$. Just switch every "left" to "right" in the above case.



Proof is done. As you have indicated, this is intuitively so obvious!



Now let us prove a proposition that is slightly more general than what you want to prove.
(Right child insertion) Let $T$ be a BST with node $Y$ that does not have a right child. Let node $X$ be a new node larger than $Y$ such that any node that is larger than $Y$, if exists, must be larger than $X$. Then node $X$, if inserted, will be inserted as the right child of $Y$.

Proof. Let $T'$ be the binary tree obtained from $T$ when $X$ is added as the right child of $Y$. Compare the inorder traversal on $T$ and that on $T'$. By the definition of inorder traversal, the difference between them can happen only when each traverses the subtree rooted at $Y$. 

Both traversals will visit the same left subtree of $Y$ first. Then both will visit $Y$.  When it is time to visit the right subtree of $Y$, the former traversal has nothing to do since $Y$ has no right child in $T$ while the latter traversal will visit one more node, $X$ since $Y$ has $X$ as its right child in $T'$.

Since the nodes in $T$ is visited in its natural order by the former traversal, the nodes in $T'$ is visited in its natural order by the latter traversal as $X$ is the very next node that is larger than $Y$. By definition, $T'$ is a BST. According to the above uniqueness lemma, $X$ must be inserted as the right child of $Y$. Q.E.D.