Regarding the motivation for your question (sequential compactness):
First, commenting on your specific question: a topological vector space over the real or complex numbers is never sequentially compact. Only bounded closed subset are possibly (sequentially) compact. Your attempt of proof is, therefore, only relevant if you restrict to bounded sequences (which applies to your example but should be made an explicit assumption).
As a general remark: it is known that compactness and sequential compactness are equivalent in metric spaces, hence in subsets of normed vector spaces (like $C^0$).
It is also known (though a bit more involved when it comes to proving it) that a normed (real or complex) vector space has the so called 'Heine Borel' property (which says that a subset is compact if and only if it is closed and bounded) if and only if it is of finite dimension.
Edit: the following sentence is obviouly not correct (see the comment of t.b. -- thanks for pointing this out), but is a nice illustration of a premature und uncautios conclusion, so I don't delete it ;-): Since $C^0$ is not finite dimenional, it follows that bounded closed sets of $C^0$ are (sequentially) compact only if they are contained in a finite dimensional subspace.
The following remains true, though: sets like the closed unit ball (and therefore any set containing an open ball) are not (sequentially) compact in $C^0$ and, in general, bounded sequences need not have a converquent subsequence, and this applies to every infinite dimensional normed vector space.