Counterexample in the general case: Take R=$\prod_{\mathbb N} \mathbb F_2$, I=$\sum_{\mathbb N} \mathbb F_2$ and C the cokernel of the inclusion.
Now I claim three things:1.) R has dimension zero
Proof: Every element of R is an idempotent and every prime ideal in R has to contain exactly one of e or 1-e for every idempotent e in R. If we had a chain of prime ideals, the larger one would nescessarily have to contain e and 1-e for one such idempotent and so couldn't exist.
2.) The localisation of R at every maximal ideal is a field.
Proof: It is a zero-dimensional local ring by 1.) and reduced since R doesn't contain nilpotent elements. Thus this localisation is a field.
3.) I is not a direct summand of R.
Proof: Direct Summands correspond to idempotents in R. Since every element in R is idempotent we need to analyze all principal ideals. If an element has only finitely many non-zero entries the ideal created by it has only finitely many elements, thus can't be I.If on the other hand it contains infinitely many non-zero entries, the ideal created by it has uncountably many elements.Since I contains countably many elements, it can't be a direct summand of R.
This establishes the counterexample, since the short exact sequence 0→I→R→C→0 splits in every localisation at a maximal ideal.
