How to prove a subspace? Let X be a Banach space and let M be a subset of X. Then M is itself
a Banach space (using the norm from X) if and only if M is closed subspace.

How to prove (=>) to show that M is closed subspace?

Question:If I have M is subset of X then I have M is a subspace???? If this statement is true, I can prove it.

Thanks. :-)
Suppose $M$ is closed. You have to show $M$ is complete. Let $x_n$ be a Cauchy Sequence in $M$. Since $X$ is a Banach space, $x_n$ converges to some $x\in X$. Since $M$ is closed, $x\in M$. So $M$ is complete.

Suppose $M$ is a Banach space. You have to show $M$ is closed in $X$. Since $X$ is a normed space it suffices to show that, if $x_n$ is a sequence in $M$ which converges in $X$ to some $x\in X$, then if follows that $x\in M$. So suppose $x_n$ is convergent to $x$. Then $x_n$ is a Cauchy Sequence in $X$, but since $M$ carries the same norm, in $M$ as well. Since $M$ is a Banach space, $x_n$ converges in $M$ to $x^\prime \in M$. Since $M\subset X$ with the induced topology, $x_n\rightarrow x^\prime$ in $X$. By uniqueness of the limit, $x = x^\prime$, so $M$ is closed.
To $M$ be a subspace you need that $\forall x,y,\in M $ and $\forall \alpha$ scalar 

$x+y\in M$ 
and
$\alpha x \in M$.

But, it doesn't mean that $M$ is a Banach space. To be a Banach space one needs that $M$ be a complete space, that's why one requires that $M$ be closed.

To show that $M$ is closed you need to show that every Cauchy sequence converges to a point in $M$.