Vector space over an infinite field which is a finite union of subspaces 
Let $V=\bigcup_{i=1}^n W_i$ where $W_i$'s are subspaces of a vector space $V$ over an infinite field $F$. Show that $V=W_r$ for some $1 \leq r \leq n$.

I know the result "Let $W_1 \cup W_2$ is a subspace of a vector space $V$ iff $W_1 \subseteq W_2$ or $W_2 \subseteq W_1$."
Now can I extend this to some $n$ subspaces.
I have some answers here
& here.
So before someone put it as a duplicate I want to mention that I want a proof of this problem using basic facts which we use in proving the mentioned result.
 A: The general result  is that if $\lvert K\rvert\ge n$, $V$ cannot be the union of $n$ proper subspaces (Avoidance lemma for vector spaces).
We'll prove that if $V=\displaystyle\bigcup_{i=1}^n W_i$ and the $W_i$s are proper subspaces of $V$, then $\;\lvert K\rvert\le n-1$.
We can suppose no subspace is contained in the union of the others. 
Pick $u\in W_1\smallsetminus\displaystyle\bigcup_{i\neq1} W_i$, and $v\notin W_1$. The set $v+Ku$ is disjoint from $W_1$, and it intersects each $W_i\enspace(i>1)$ in at most $1$ point (otherwise $u$ would belong to $W_i$). As this set is in bijection with $K$, there results that $K$ has at most $n-1$ elements.
A: $V$  is  a  vector  space  over  an  infinite  field.
If $\ \ $   $\cup_{i=1}^{n} W_{i}=V$  is  a vector space  itself  then  from  what  you  already  know we  can  write $$W_{1}\subseteq W_{2}\subseteq W_{3}........\subseteq W_{n}$$  possibly  with  a  re-ordering  of  the  subspaces  $W_{i}$'s .  
Then  what  do  we  see  here?  $W_{n}$  containing  the  other  subspaces  of  $V$. So  intuitively  we  could  suspect  that  $V=W_{n}$  might  be  the  thing .
Let's  prove  it. $$W_{n}\subseteq V$$  is known.  For  the  reverse  inclusion ; if possible  let  $$V\neq W_{n}$$ let $w$  be  a  vector such  that $$w\in V \ \  but\ \    w\notin W_{n}$$  . Now  $$\cup_{i=1}^{n} W_{i}=V$$  implies  $w$  is  in  $W_{i}$  for  some  $i=1,2...,n$. But  $$W_{i}\subseteq W_{n}$$ for  all  $i=1,2,....n$ .  Hence $$w\in W_{n}$$ . A  contradiction.  Thus  $$V\subseteq W_{n}$$  must  be  true. Hence  proved  $$V=W_{n}$$ 
