I would do this problem as follows. Let x ∈ (1, 5) and let e = min(x - 1, 5 - x) > 0. Then there exists an n such that 1/n < e. Therefore, x ∈ [1 + e, 5 - e] ⊆ [1 + 1/n, 5 - 1/n] ⊆ ∪_{k=1}^{∞}[1 + 1/k, 5 - 1/k]. The converse inclusion is obvious because [1 + 1/k, 5 - 1/k] ⊆ (1, 5) for every k.