Subcomplex is a closed subspace

Suppose A ⊆ |L| ⊆ |K|.
1. If T_α∈K, |L| ∩ T_α is a finite union of faces of T_α hence is closed in T_α. So |L| is closed in |K|.
2. If A is closed in the subspace topology of |L| then it is closed in |K| so is closed in |L|.
3. Suppose A is closed in |L|. If T_α∈K, A ∩ T_α = A ∩ (T_α ∩ |L|). T_α ∩ |L| is a finite union of faces f_i, i = 1, 2, ..., n, of T_α which are also in L. So A ∩ T_α is a finite union of A ∩ f_i , i = 1, 2, ..., n, each of which is closed in f_i and hence in T_α. So A is closed in |K|. Hence A is closed in the subspace topology. QED