Is the unit sphere in an infinite dimensional Hilbert space closed? Is a unit sphere in an infinite dimensional hilbert space closed. By the triangle inequality it is clear that the all the limit points of the sphere are inside the closed unit ball. But I cannot somehow show that limits of sequence in the unit sphere lie on the unit sphere. 
Or maybe I am not thinking straight. How could I show it?

Converting comments to answers:  This is true in any normed space.  (In fact, it is true in every metric space that spheres are closed.)
A variant of the triangle inequality is the fact that
$$\bigg| \|x\| - \|y\| \bigg| \le \|x-y\|. \tag{1}$$
To prove this, use the triangle inequality to see that
$$\|x\| = \|x - y + y\| \le \|x-y\| + \|y\|$$
which shows $\|x\| - \|y\| \le \|x-y\|$.  Similarly,
$$\|y\| = \|y - x + x\| \le \|y-x\| + \|x\| = \|x-y\| + \|x\|$$
which shows $\|y\| - \|x\| \le \|x-y\|$.  Combining these gives (1).
Now suppose $\|x_n\| = 1$ for all $n$ and $x_n \to x$.  For every $n$ we have
$$\bigg| 1 - \|x\| \bigg| = \bigg| \|x_n\| - \|x\| \bigg| \le \|x_n - x\|.$$
As $n \to \infty$, the right side converges to 0.  Hence the left side must be 0, which is to say $\|x\|= 1$.
