Prove $T$ is a BST iff for every node $x$ of $T$ that is not a leaf, the key of $x$ is larger or equal than the key of the left child of $x$
The second direction is not always true, in a BST all elements in the right subtree of a node should have a key bigger than its key and all elements in the left subtree of a node should have a key less than its key. for example, consider this binary tree:


     4
   /   \
  2     6
 / \   / \
1   9 3   7


it has the property that for every node x of $T$ that is not a leaf, the key of x is larger or equal than the key of the left child and less or equal than the key of the right child, but it's not a BST since 3 is in the right subtree of root but its key is less than the key of root.