(1) Denote the common difference of the arithmetic sequence {an} as d. From S3 = 15 and a3 + a8 = 2a5 + 2, we obtain:
3a1 + 3d = 15,
2a1 + 9d = 2(a1 + 4d) + 2.

Solving these equations, we get d = 2 and a1 = 3.

Hence, an = a1 + (n - 1)d = 3 + 2(n - 1) = 2n + 1.

(2) Proof:
$$\frac{1}{S_n} = \frac{1}{3n + \frac{1}{2}n(n - 1) \cdot 2} = \frac{1}{n(n + 2)} = \frac{1}{2} \left( \frac{1}{n} - \frac{1}{n + 2} \right)$$

The sum of the first n terms, Tn, is given by:
Tn = $$\frac{1}{2}$$ (1 - $$\frac{1}{3}$$ + $$\frac{1}{2}$$ - $$\frac{1}{4}$$ + $$\frac{1}{3}$$ - $$\frac{1}{5}$$ + ... + $$\frac{1}{n - 1}$$ - $$\frac{1}{n + 1}$$ + $$\frac{1}{n}$$ - $$\frac{1}{n + 2}$$)
= $$\frac{1}{2}$$ (1 + $$\frac{1}{2}$$ - $$\frac{1}{n + 1}$$ - $$\frac{1}{n + 2}$$)
= $$\frac{3}{4}$$ - $$\frac{2n + 3}{2(n + 1)(n + 2)}$$.

Since $$\frac{2n + 3}{2(n + 1)(n + 2)} > 0$$, we have Tn < $$\boxed{\frac{3}{4}}$$.