Suppose $\sum X_{n}$ converges a.s. and $|X_n| \leq K$. Define $T:=\inf\{\tau:|M_{\tau}|>c\}$ where $M_{\tau}:=\sum_{k=1}^{n}X_{k}$.
Then why is $P(T=\infty)>0$ for some $c$?
Since $\sum X_{n}$ converges a.s., the partial sums of $\sum X_{k}$ are a.s. bounded, and it must be the case that for some $c$, $P(T=\infty)>0$.
I don't understand what the above line is saying.
Set
$$Y(\omega) := \sup_{n \in \mathbb{N}}|M_n(\omega)| = \sup_{n \in \mathbb{N}} \left| \sum_{j=1}^n X_j(\omega) \right|.$$
By assumption, $Y$ takes (almost surely) values in $\mathbb{R}$. Since
$$1= \mathbb{P}(Y \in \mathbb{R}) = \lim_{k \to \infty} \mathbb{P}(|Y| \leq k)$$
we can choose $k>0$ such that $\mathbb{P}(|Y| \leq k)>0$. If we choose $c:=k$, then
$$\{|Y| \leq k\} = \{\forall n \in \mathbb{N}: |M_n| \leq k\} \subseteq \{T=\infty\}$$
and so
$$\mathbb{P}(T=\infty) \geq \mathbb{P}(|Y| \leq k)>0.$$