I will assume that $\alpha_i\in(0,1)$.
Note that $0\le\prod_{i=0}^k (1-\alpha_i) \le \prod_{i=0}^k \frac1{1+\alpha_i} = \frac1{\prod_{i=0}^k(1+\alpha_i)} \le \frac 1{1+\sum_{i=0}^k \alpha_i}$
For $k\to \infty$ the RHS tends to zero, so you get $\lim\limits_{k\to\infty} S_k=0$.
We have used $1-x\le \frac1{1+x}$, which follows from $(1-x)(1+x)=1-x^2\le 1$.
For the second part, let us try to use this result: $\liminf_{n\to\infty}\frac{c_{n+1}}{c_n}\leq\liminf_{n\to\infty}\sqrt[n]{c_n} \le \limsup_{n\to\infty}\sqrt[n]{c_n}\leq\limsup_{n\to\infty}\frac{c_{n+1}}{c_n}$ which is true for any positive sequence $(c_n)$, see e.g. this answer and this question.
If we apply this to the sequence $(S_k)$, we get $\liminf_{k\to\infty} (1-\alpha_{k+1}) \le \liminf_{k\to\infty} \sqrt[k]{S_k} \le \limsup_{k\to\infty} \sqrt[k]{S_k} \le \limsup_{k\to\infty} (1-\alpha_{k+1}),$ which implies $\lim\limits_{k\to\infty} \sqrt[k]{S_k}=1$.