Convergence a.s. $\Longleftrightarrow$ Convergence in P 
Let $\{Y_n\}'s$ be independent r.v.'s, $S_n = \sum^n_{i=1}Y_i$,
a. Prove that for arbitrary $\epsilon >0$, $$P\left(\sup_{\{k\geq1\}}|S_k|>4 \epsilon\right) \leq 4\sup_{\{k\geq1\}}P(|S_k|>\epsilon)$$ 
  and for each integer $n$, $\epsilon >0$, 
  $$P\left(\sup_{\{n\geq k\geq1\}}|S_k|>4 \epsilon\right) \leq 4\sup_{\{n\geq 
   k\geq1\}}P(|S_k|>\epsilon).$$
b. Use (a) to prove $\sum^{\infty}_{i=1}Y_i$ convergence in probability implies it convergence almost surely.

I am very confused about how to solve this question, by my intuition, convergence in probability dose not implies convergence almost surely. 
 A: The second inequality is a direct consequence of Etemadi's inequality and you can find a proof, for instance, here. To derive the first inequality from the second we note that
$$\left\{ \sup_{k \geq 1} |S_k|>4 \epsilon \right\} = \bigcup_{n \geq 1} A_n$$
for
$$A_n := \left\{ \sup_{1 \leq k \leq n} |S_k|>4 \epsilon \right\}.$$
Since the sequence $(A_n)_{n \in \mathbb{N}}$ is increasing, the continuity of the measure $\mathbb{P}$ gives
$$\mathbb{P}(A) = \lim_{n \to \infty} \mathbb{P}(A_n).$$
Using the second inequality for $\mathbb{P}(A_n)$, the first inequality follows.
Regarding part (b): The key tool is the following characterization of almost sure convergence.

A sequence $(S_n)_{n \in \mathbb{N}}$ converges almost surely if, and only if, for any $\epsilon>0$
  $$\lim_{m \to \infty} \mathbb{P} \left( \sup_{n \geq m} |S_m-S_n|> \epsilon \right)=0.$$

If $(S_n)_{n \in \mathbb{N}}$ converges in probability, then we can choose for any $\epsilon>0$ and $\delta>0$ some $N \in \mathbb{N}$ such that
$$\sup_{n \geq m} \mathbb{P}(|S_n-S_m|>\epsilon) \leq \delta$$
for all $m \geq N$. Applying the first inequality (for the sequence $\tilde{S}_n := S_n-S_m= \sum_{i=m+1}^n Y_i $) for fixed $m \geq N$, we get
$$\mathbb{P} \left( \sup_{n \geq m} |S_n-S_m|>4\epsilon \right) \leq 4 \sup_{m \geq n} \mathbb{P}(|S_n-S_m|>\epsilon) \leq 4\delta.$$
As $\delta>0$ is arbritrary, this shows
$$\lim_{m \to \infty} \mathbb{P} \left( \sup_{n \geq m} |S_m-S_n|> 4\epsilon \right)=0,$$
and hence almost sure convergence.
