Three space property I want to show that Finite dimensionality is a three space property. Let $X$ be a normed linear space and let $Y$ be a closed subspace of $X$. If $Y$ and $X/Y$ are finite dimensional spaces, then I want to show $X$ to be finite dimensional. 
Let $B$ be a finite basis of $Y$. Then it can be extended to a basis $B^{\prime}$ of $X$. But how to show $B^{\prime}$  to be finite? Any hint will be appreciated.

Let $x_1,\ldots,x_n\in X$ be such that $(x_1+Y,\ldots,x_n+Y)$ is a basis of $X/Y$ and let $(y_1,\ldots,y_m)$ be a basis of $Y$. If $x\in X$, then there are scalars $\alpha_1,\ldots,\alpha_n$ such that$$x+Y=\alpha_1(x_1+Y)+\cdots+\alpha_n(x_n+Y).$$So, $x-\sum_{k=1}^n\alpha_kx_k\in Y$ and therefore there are scalars $\beta_1,\ldots,\beta_m$ such that$$x-\sum_{k=1}^n\alpha_kx_k=\sum_{l=1}^m\beta_ly_l.$$Therefore, $X$ is spanned by $\bigl\{x_k+y_l\,|\,(k,l)\in\{1,\ldots,n\}\times\{1,\ldots,m\}\bigr\}$. So, $X$ is finite-dimensional.
