Linear subspace and dimension If $Y$ is a proper linear subspace of a finite dimensional linear space $X$, I want to show that $Y$ is also finite dimensional and $\dim(Y)< \dim(X)$.

Hint:
If $\,Y=\{0\}\,$ there's nothing to prove, otherwise take $\,0\neq y_1\in Y\,$ . If $\,\operatorname{Span}\{y_1\}=Y\,$ we're done otherwise there exists $\,Y\ni y_2\notin \operatorname{Span}\{y_1\}\,$ . 
Then $\,\{y_1,y_2\}\,$ is linearly independent (why?), so if $\,\operatorname{Span}\{y_1,y_2\}=Y\,$ we done, otherwise there exists $\,Y\ni y_3\notin \operatorname{Span}\{y_1,y_2\}\,$ , so then $\,\{y_1,y_2,y_3\}\,$ is lin. independent (again, why?) ...
Continue as above, but note all this happens within $\,X\,$ and here a linearly independent set has only a finite number of elements...complete and end the argument.
