Is the infinite-dimensional unit sphere compact?

Suppose that the unit sphere $S_X$ of $(X, \| . \|_X)$ is compact. Then the unit ball of $X$ is the image of the compact set $[0,1] \times S_X$ by the continuous map $(t, v) \mapsto tv$, and hence is compact.

Sounds right to me. The sequence of points $(1,0,0,\ldots), (0, 1, 0, \ldots), (0, 0, 1, \ldots), \ldots$ is a sequence of points on the sphere that has no convergent subsequence, because the distance between any two of the points is $\sqrt{2}$.

Hint: In a metric space, sequential compactness and compactness are equivalent. Now consider a sequence consisting of unit length basis vectors.