Short proof of Hall's theorem Studying the proof of Hall's theorem in my book I started to wonder if there is a shorter way to prove it. Following is an attempt that I think works but (being short) makes me wonder if I made a mistake. Can someone double check?

For a collection of sets $S_1,\ldots,S_m$ we say that they have a
  collection of distinct represenatives if there exists distinct
  elements $x_1,\ldots,x_m$  such that $x_i \in S_i$ for $i = 1,\ldots, m.$ We say that our  collection of sets satisfies Hall's criterion if $$ \left |\bigcup_{i \in I} S_i \right | \geq |I| \quad \mbox{for all } I \subseteq \{1,\ldots,m\}.$$

Claim. If $S_1,\ldots,S_m$ satisfy Hall's criterion then they have a set of distinct representatives. 
Proof. We prove the claim by induction on $m.$ For $m=1$ the claim is obvious so suppose the claim holds for all $m < k$ for some $k$ and let $S_1,\ldots,S_m$ be a collection of sets satisfying Hall's criterion. Let $\{x_1,\ldots,x_{m-1}\}$ be the set of distinct representatives for $S_1,\ldots,S_{m-1}$ guaranteed by the indcution hypothesis. If $\{x_1,\ldots,x_{m-1}\} \ne S_m$ then we are done. If $\{x_1,\ldots,x_{m-1}\} = S_m$ that then $ \left |\bigcup_{i=1}^{m-1} S_i \right | > m-1$ since otherwise $S_1,\ldots,S_m$ do not satisfy Hall's criterion. But this implies that for some $i$ there exist a $x_i' \in S_i$ such that $x_i' \not \in S_m$ and hence we can take $\{x_1,\ldots,x_{i-1}, x_i',x_{i+1},\ldots,x_{m-1},x_i\}$ as our set of distinct representatives. $\square$
Anyone happens to see a flaw in the above proof?

When you say "If $\{x_1, \ldots, x_{m-1}\} \ne S_m$ we are done", I do not see why that is the case.  It could be that $S_m$ is a proper subset of $\{x_1, \ldots, x_{m-1}\}$, in which case your subsequent argument does not work, because you do not know that $x_i \in S_m$.  
