Since the connected components of a metric space X are equivalent classes in X, X can be partitioned into the connected components of X. Let be the connected component of X containing x. For points x,y in X, the connected component containing x and the connected component containing y are either identical or disjoint. Thus, each point belongs to exactly one connected component which is the largest connected subset of X containing x.
Assume that is a connected proper subset of . By lemma 1, is a connected subset containing x, contradicting that is the largest connected subset of X containing x. Thus, .
We conclude that , is closed.