Let $G$ be a bipartite graph with parts $A$ and $B$. Show that if $G$ has a matching that covers every vertex of $A$, then for every $A'\subseteq A$, we have $|N(A')|\geq|A'|$, where $N(A')$ denotes the set of neighbours of $A'$.
This sseems pretty obvious but I can't put it into correct words. If $G$ has a matching that covers every vertex of $A$ then the degree of all vertices of $A$ is going to be at least $1$ or else the matching wouldn't contain all vertices of $A$. So the number of neighbours of $A'$ is going to be at least the number of vertices of $A'$.
I don't know how to be rigorous with this. Also if anything that I have said is incorrect in any way, then please say.
Ok here is my new answer: Let $G=(V,E)$. Consider A'. Since we have a matching that covers all vertices of A, then this matching clearly covers all vertices of A'. The degree in each vertex of A' is going to be at least 1 - otherwise there cannot be a matching that covers all vertices in A'. Further, there cannot be any edge that share vertices in A' otherwise G would not be a bipartite graph. So for edge $uv \in E$, either u is in A' and v is in B or vise versa. This means edges that come from vertices in A' is going to be joined to vertices in B. Since we said the deg of each vertex in A' is at least 1, then there is clearly going to be at least |A'| edges that are incident to vertices in A'.
Since we have a matching that covers all vertices in A', and since each vertex degree is at least 1 in A', each vertex is going surely going to have at least $1$ neighbour that is different to the other neighbours that the other vertices in A' have. If it didn't then there would be an edge that shares the same vertex in A' meaning there would be no matching (that covers all vertices of A'). So this means there is at least going to |A'| neighbours in vertices of A' that are different to each other. This means $|N(A')|\geq |A'|$.
Now what do you think of it now? Again please point out ANYTHING that is wrong with it or can be improved or can be justified more.