Totally bounded and closed subset of complete metric space => compact

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I've read that a totally bounded and closed set $A \subseteq E$ is compact if $E$ is a complete metric space. There was not a direct proof they only referenced a theorem of another book. The theorem in that book states, that a complete space $E$ is compact if and only if $E$ is complete and totally bounded. But that does not really explain why totally bounded and closed subsets of $E$ are compact or am I missing something?

Thanks for your help.