A company has a number of stations made up of workers. Each station has at least $5$ workers and no worker can be on $6$ different stations. Can the owner assign one worker from each station to be the manager of that station in such a way that a worker is the manager of at most one station?
I am thinking that we can use Hall's marriage theorem to solve this problem.
Let $A$ be a set of vertices denoting the stations and $B$ a set of vertices denoting the workers, then we simply need to show that for any set of workers $W\subseteq A$, we can always have that $|N(W)|\geq|W|$.
Now, let the vertices $w_i's$ denote the workers and $s_i's$ the stations. We must have that $\deg(w_i) \leq 5$ and $\deg(s_i)\geq 5$.
I am not sure if my reasoning is correct as i'm having problems formalizing it, but I am inclined to think that this is true, since if we consider the neighbours of $5$ stations, we require at least $5$ workers, and to increase the number of stations in such a way that there are less workers would mean that some worker was on more than $5$ stations.
You're on the right track, but showing that $|N(W)| \ge |W|$ for any $W \subseteq A$ is the condition for having an $A$-perfect matching: a matching of workers to stations that means every worker gets to be manager.
This isn't what the question is asking, and is also easy to rule out by a counter-example: if Alice and Bob work together at Station X, and don't work anywhere else, then at most one of them can be a manager.
Instead, we want to make sure that every station is assigned a worker, and for this we need to get a $B$-perfect matching. So you want to check that for every set of stations $S \subseteq B$, $|N(S)| \ge |S|$.
This can be shown by counting the edges from $S$ to $N(S)$. At least many edges are there coming out of a set $S \subseteq B$? How many workers are required to absorb that many edges?