<p>We characterize conformally removable sets in the plane with the aid of the recent developments in the theory of metric surfaces. We prove that a compact set in the plane is <i>S</i>-removable if and only if there exists a quasiconformal map from the plane onto a metric surface that maps the given set to a set of linear measure zero. The statement fails if we consider maps into the plane rather than metric surfaces. Moreover, we prove that a set is <i>S</i>-removable (resp. <i>CH</i>-removable) if and only if every homeomorphism from the plane onto a metric surface (resp. reciprocal metric surface) that is quasiconformal in the complement of the given set is quasiconformal everywhere.</p>