I am asked to prove that $$ X=\left\{x=(x_n)_{n\in \mathbb N} :\sum_{n\in \mathbb N} |x_n|^{p(n)}<\infty \right\}$$
with $p(n)>0$ is a linear space iff $\sup_n p(n)<\infty $
The direction $(\Leftarrow)$, that is, that $\sup_n p(n)<\infty$ implies $X$ is linear is not that difficult. I do not know how to prove the other direction $(\Rightarrow)$.
I'm going to use $X_p$ to denote the set above. Suppose $X_p$ is an $\newcommand{\RR}{\mathbb{R}}\RR$-vector space, or at least a $\Bbb{Q}$-vector space. Suppose for contradiction that $p$ is unbounded. By unboundedness, choose a subsequence $n_k$ such that $p(n_k) \ge k$. Now consider $$x(n) = \begin{cases}\frac{1}{2} & \text{if $n=n_k$ for some $k$,}\\ 0 & \text{otherwise.}\end{cases}$$ By construction, $$\sum_{n=0}^\infty |x(n)|^{p(n)} = \sum_{k=0}^\infty \left(\frac{1}{2}\right)^{p(n_k)} \le \sum_{k=0}^\infty \left(\frac{1}{2}\right)^{k}=2, $$ so $x\in X_p$. However $2x$ is not in $p$, since $$\sum_{n=0}^\infty |2x(n)|^{p(n)} = \sum_{k=0}^\infty \left(1\right)^{p(n_k)} = \sum_{k=0}^\infty 1 = \infty. $$ Contradiction. Thus if $X_p$ is a $\Bbb{Q}$-vector space, $p$ is bounded.