Let $X = \{ f \in \Bbb Q^\Bbb R : \sum_{x \in \Bbb R} |f(x)| < \infty \}$. Show that $X$ is a dense subset of $\mathbb{R}^{\mathbb{R}}$ with the product topology and that no sequence in $X$ converges to the constant function $\mathbf{1} \in \mathbb{R}^\mathbb{R}$.
Let $f$ be a function in $\mathbb{R}^\mathbb{R}$ and consider any basic open subset $O$ of $f$. By the product topology $O$ takes the form $$O=\prod_{x \in \mathbb{R}}O_x$$ where $O_x = \mathbb{R}$ for all $x \in \mathbb{R} \setminus F$ ($F$ is a finite subset of $\mathbb{R}$) and $O_x \subset \mathbb{R}$ are open for $x \in F$.
Since $\mathbb{Q}$ is dense in $\mathbb{R}$ for each $x \in F$ there exists $q_x \in O_x$ and thus we can define $$g : \mathbb{R} \to \mathbb{Q}, \ g(x)= \begin{cases} q_x &\ \text{if } x \in F \\ 0 &\ \text{if } x \notin F\end{cases}.$$
This is now a function that is in $O$ and it's in $X$ as $$\sum_{x \in \Bbb R} |g(x)| < \infty$$ since we're summing over a finite number of rationals. Thus $O \cap X \ne \emptyset$ and $X$ is dense.
I have some issues with the second part of the problem. If $(f_n)_n$ is a sequence in $X$ such that $f_n \to \mathbf{1}$ pointwise, then $$\forall x \in \mathbb{R} : \forall \varepsilon >0 : \exists n_0 \in \mathbb{N} : |f_n(x)-1|< \varepsilon.$$
I have a feeling that I should be able to contradict the sum $\sum_{x \in \Bbb R} |f(x)| < \infty$ somehow just from this convergence criterion, but I cannot figure out how?
Note that if $\sum_{x \in \mathbb{R}} |f(x)| < \infty$ then $\{x: f(x) \neq 0\}$ is countable.
To see this, write $\{x: f(x) \neq 0\} = \bigcup_{n \ge 1} \{x: |f(x)| > \frac{1}{n}\}$. If the left hand side is uncountable then at least one set in the union on the right hand side is uncountable (and in particular infinite) so that the sum must also be infinite.
This means that given a sequence $f_n$ in $X$, $\{x: \exists n \text{ such that } f_n(x) \neq 0\}$ is also countable. Since $\mathbb{R}$ is uncountable, $f_n$ cannot converge pointwise to $1$.