Let $X_n$ be a sequence of independent random variables, with $\mathbb E(X_n)=0$ and $|X_n|<K, \forall n,\omega$, then:
P$(\sup_{n\in\mathbb N}|\sum_{k=1}^{n}X_k|<\infty)>0 \iff$ P$(\sum_{k=1}^{\infty}X_k$ exists in $\mathbb R)=1$
This can be also found in David Williams, probability with martingales (Remark in p.113). But this is not proven there.
New question deleted. It can be found here.
"$\Leftarrow$" is obvious. For "$\Rightarrow$" we set $\sigma_k^2 = \mathbb{E}(X_k^2)$, $$M_n := \sum_{k=1}^n X_k \qquad \quad A_n := \sum_{k=1}^n \sigma_k^2.$$ Using the independence of the random variables, it is not difficult to see that $(M_n^2-A_n)_{n \geq 0}$ is a martingale (with respect to the canonical filtration). For $r>0$ we define a stopping time $\tau$ by
$$\tau := \inf\{n; |M_n| > r\}.$$
Since, by assumption,
$$\mathbb{P} \left( \sup_{n \in \mathbb{N}} \left| \sum_{k=1}^n X_k \right|<\infty \right)>0$$
we have $$\mathbb{P}(\tau=\infty)>0 \tag{1}$$ for $r$ sufficiently large.
By the optional stopping theorem,
$$\mathbb{E}(M_{n \wedge \tau}^2-A_{n \wedge \tau})=0;$$
hence,
$$\mathbb{E}A_{n \wedge \tau} = \mathbb{E}(M_{n \wedge \tau}^2) \leq (K+r)^2. \tag{2}$$
By $(1)$, we have $\tau(\omega)= \infty$ for each $\omega \in A$ for $r$ sufficiently large and some set $A \in \mathcal{A}$ such that $\mathbb{P}(A)>0$. Therefore, $(2)$ implies by the monotone convergence theorem
$$\sum_{k=1}^{\infty} \sigma_k^2 < \infty.$$
This means that
$$\mathbb{E} \left[ \left( \sum_{j=1}^{\infty} X_j \right)^2 \right] = \sum_{j=1}^{\infty} \mathbb{E}(X_j^2) = \sum_{j=1}^{\infty} \sigma_j^2 <\infty.$$
In particular, $\sum_{j=1}^{\infty} X_j<\infty$ almost surely.