It is absolutely intuitive that it has to be the case, but besides drawing (an example) a Bipartite Graph for |X|=3 and |Y|=3 and showing that for |X|=3 and |Y|=3, |X|=3 and |Y|=4 and |X|=4 and |Y|=4 with the minimal degree always being equal to 3 that its matching will always consist of at least 3 edges (and then generalising it for n instead of 3), I don't have an idea for a more "formal proof".
Such a "proof by example" is simply not satisfying to me
I think you can just take one vertex from X and assign it to one of its neighbors in Y. Now take another vertex from X and worst case scenario it will have n-1 neighbors that are not yet taken. Repeat the process and in the end you will take the nth vertex from X, which will have at least 1 neighbor that was not taken. You are done here.