My doubt is to prove just what is underlined in red. I tried two ways:
first
$ g_k = | f_{n_2} - f_{n_1} | + ... + | f_{n_{k+1}} - f_{n_{k}} | $ , then $ g_k^p = (| f_{n_2} - f_{n_1} | + ... + | f_{n_{k+1}} - f_{n_{k}} |)^p $
by Minkowski we have
$(\int_{X} g_k^pd\mu)^{1/p} \leq (\int_{X} | f_{n_2} - f_{n_1} |^p d\mu)^{1/p} +... + (\int_{X} | f_{n_{k+1}} - f_{n_{k}} |^p d\mu)^{1/p} $ $ \hspace{1.6cm} \leq (\int_{X} \frac{1}{2^p} d\mu)^{1/p} +... + (\int_{X} \frac{1}{2^{kp}} d\mu)^{1/p} $
$ \hspace{1.6cm} \leq \frac{1}{2} \mu(X)^{1/p} +... + \frac{1}{2^{k}} \mu(X)^{1/p} $
but to use the latter inequality it is necessary that $ \mu(X) < \infty $. which is not in the hypothesis.
the second way I thought it was
for each $i \in \{1, ..., k\}$
by Minkowski we have
$ (\int_{X} | f_{n_{i+1}} - f_{n_{i}} |^p d\mu)^{1/p} \leq (\int_{X} | f_{n_{k+1}} |^p d\mu)^{1/p} + (\int_{X} |f_{n_{k}} |^p d\mu)^{1/p} $
but, here I can not use the inequality $|| f_{n_{i+1}} - f_{n_{i}} || < \frac{1}{2^{i}} $ to complete something useful.
Let $h_{i}=|f_{n_{i+1}}-f_{n_{i}}|$, then $\|h_{i}\|_{p}<2^{-i}$ and $g_{k}=h_{1}+\cdots+h_{k}$, so $\|g_{k}\|_{p}\leq\|h_{1}\|_{p}+\cdots+\|h_{k}\|_{p}<2^{-1}+\cdots+2^{-k}<2^{-1}+\cdots=1$.