Could someone please check my work

If S is a compact subset of R and T is a closed subset of S , then T is compact. 1. Prove this using the definition of compactness. Assume S is a compact subset of IR and T Q S, and T is a closed. Let AT = {An : n E N} be any open cover for T . If AT is also a cover for S , then since S is compact, by definition 3.5.1 for compactness, => there exists a finite subcover of AT, {An .. 1...An _ k}, that covers S . => Since T g S , then {An _ 1...An _ Is} also covers T . Conversely, if AT is not an open cover of S and merely covers T , since S is compact, then by definition 3.5.1 for compactness, there must be an open cover for S . => Thus, add an OPEN set to open set, AT in order to cover 5. Since T is closed, then by Theorem 3.4.7(b), {R T} is open. ' Since S is compact and T C_: S , then by definition 3.5.1 for compactness and the definition of subset, SinceTQSQR={RT}UT:{RT}UAT, i 2? {RT}UATisan open coverofSandT. So adding open set, R T to open set, AT will produce an open cover of S as desired. ' However, since R T contains no points in T, AT is a sufficient open cover of T . i Since S is compact and T Q S , then by definition 3.5.1 for compactness and the definition of subset, the open covers of T and S must be reducible to finite subcovers of T and S . SinceTQ SE; {RT}U{An_l,...,An_k} => {An _ 1, ..., An _ k} U (R T) is a finite subcover of Sand T. But again, since IR - T contains no points in T ,{An _ 1...An __ k} is a sufficient i , of T. >5 In either case, T is covered by a finite number of open sets, {A" _. 1...An __ k} . Therefore, T is compact