Simple Proof of Obvious Fact. Have you any idea of how to proof the following as simpler as possible?
Let $M$ be family of $n$ finite sets:
$$M = \{X_1, X_2, ..., X_n\}$$
Then 
$$\bigr(\forall k = 1, 2, ..., n : |X_{i_{1}}\cup X_{i_{2}}\cup\cdots\cup X_{i_{k}}|\geq k\bigl)\Rightarrow\bigl(\exists \chi_1,\chi_2,...\chi_n : \chi_i\neq\chi_j,\,\chi_i\in X_i\bigr)$$
In other words
If union of any k sets has at least k elements then there exist n different elements which represent whole family M.
 A: Your obvious fact is a harder part of a generalization of Hall’s Marriage Theorem, proved in 1935 and it probably has no very simple proof (at least I was not told about it in my graph theory lecture course). 
A: This is one of the formulations of the Hall's theorem, and yes, there is a simple proof. It proceeds by induction and starts by considering two cases:


*

*$$\bigr(\forall k \in \{1, 2, ..., n-1\} : |X_{i_{1}}\cup X_{i_{2}}\cup\cdots\cup X_{i_{k}}|> k\bigl)$$

*$$\bigr(\exists k \in \{1, 2, ..., n-1\} : |X_{i_{1}}\cup X_{i_{2}}\cup\cdots\cup X_{i_{k}}|= k\bigl)$$


The first case can be solved by arbitrary assignment and induction hypothesis. The second case splits the graph into two subsets that can be solved independently and again applies induction hypothesis. E.g., see this.
I hope this helps $\ddot\smile$
A: This proof was politely provided by @dtldarek (see comments). It is connected with graph theory. Here I just reformulate it without using graphs.  
Let $M$ be a family of $n$ sets as above. So
$$M=\{X_1, ..., X_n\}.$$
Also we will write $A\in\mathfrak{Hall}$ if $A$ satisfies condition that any union of $k$ sets from $A$ contains at least $k$ elements. We have the following 
Theorem. $M \in \mathfrak{Hall}$ if and only if there exist $n$ different elements which represent all family $M$.
Proof (Necessity)$\rhd$ Suppose there exist $n$ different elements which represent all family $M$. Let's call them $\chi_1, \chi_2, ..., \chi_n$. Without loss of generality we can write
$$\chi_i\in X_i,\, i\in\{1,2,...,n\}.$$
Therefore
$$|X_{i_1}\cup X_{i_2}\cup\cdots\cup X_{i_k}|\geq|\{\chi_{i_1}, \chi_{i_2},..., \chi_{i_k}\}|\geq k\quad \lhd$$  
(Sufficiency)$\rhd$ Let us consider two cases.


*

*Case 1. For all $k\in\{1,2,...,n-1\}$ it is fulfilled that
$$|X_{i_1}\cup X_{i_2}\cup\cdots\cup X_{i_k}|\geq k+1$$
We will prove the statement of the theorem by induction on number of sets in family. Let us denote family which contains $f < n$ sets through $M_f$. Then
The basis of induction. Obviously 
$$M_1\in\mathfrak{Hall}$$
Inductive step. Suppose that for each $f < n$ holds $M_f\in\mathfrak{Hall}.$ Then prove that for $M_n = M$ it also holds. Let us choose an arbitrary element $x_1\in X_1$ to be $\chi_1$. Now delete all set $X_1$ from $M$. So now we have $M_{n-1}$. It is clear that 
$$M_{n-1}\in\mathfrak{Hall}$$. Indeed taking some $x_1$ as $\chi_1$ we disabled only one $x$ from whole family to be another $\chi_i$. So after that we have for sets from $M_{n-1}$ 
$$|X_{i_1}\cup\cdots\cup X_{i_k}|\geq k.$$ Under inductive hypothesis we conclude that statement of the theorem is true.

*Case 2. Suppose there is $k$ so for some sets from $M$ 
$$|X_{i_1}\cup X_{i_2}\cup\cdots\cup X_{i_k}| = k.$$
Then we can divide $M$ into 2 families $M_k$ and $M_{n-k}$ so
$$M_k = \{X_{i_1}, X_{i_2}, ..., X_{i_k}\}$$
and
$$M_{n-k} = \{Y_{i_{k+1}}, Y_{i_{k+2}}, ..., Y_{i_{n}}\},$$
Where $Y_{i_{k+j}} = X_{i_{k+j}}\setminus \bigcup\limits_{t=1}^{k}X_{t}$. 
In other words in the second family we leave $n-k$ sets but subtract from them all elements of first $k$ sets. Next it is clear that
$$M_{k}\in\mathfrak{Hall}\quad\text{and}\quad M_{n-k}\in\mathfrak{Hall}.$$
For $M_k$ this is obvious. For $M_{n-k}$ suppose that there exist $l$ so
$$|Z_{l} = Y_{i_{1}} \cup Y_{i_{2}} \cup ... \cup Y_{i_{l}}| < l .$$
But let us take the union $Z_l\cup X_{1}\cup X_{2}\cdots X_{k}$. So
$$|Z_l\cup X_{1}\cup X_{2}\cdots X_{k}| < k+l.$$ Aha! Under our construction of $Y_{j}$ this cannot be so because this union is the same as
$$X_{j_1}\cup X_{j_2}\cup\cdots X_{j_l}\cup X_{1}\cup X_{2}\cdots X_{k}, \quad \text{for some}\quad j_1>j_2> ...> j_l>k$$
which has at least $l+k$ elements. Inductive hypothesis finishes proof.$\quad \lhd$
