Let $X \subseteq \mathbb{R}$, and suppose we have the following property:
For every $\varepsilon > 0$ there exists a sequence $(x_n)_{n = 1}^{\infty} \subseteq X$ such that $$ \limsup_{n \to \infty} x_n \leq \varepsilon. $$
Then there exists a sequence $(x_n)_{n=1}^{\infty} \subseteq X$ such that $$ \limsup_{n \to \infty} x_n \leq 0. $$ We can show this by using a "diagonal" argument. Let $(x_n^{(k)})_{n=1}^{\infty} \subseteq X$ have the property that $$ \limsup_{n \to \infty} x_n^{(k)} \leq k^{-1}. $$ Now, for all $k$ there exists an $N_k \in \mathbb{N}$ such that, whenever $n \geq N_k$, we have $x_n^{(k)} \leq k^{-1}$. Without loss of generality, let's assume that $N_1 < N_2 < \cdots$. Then we define the sequence $(x_n)_{n=1}^{\infty}$ as follows. For $n=1, 2, \dots, N_2$, set $x_n = x_n^{(1)}$. Given that the sequence has been defined up until $N_k$, set $x_n = x_n^{(k)}$ for all $n = N_k + 1, N_k + 2, \dots, N_{k+1}$.
It follows that for all $k \in \mathbb{N}$ there is an $N_k \in \mathbb{N}$ such that for all $n \geq N_k$ we have $x_n \leq k^{-1}$, which implies $$ \limsup_{n \to \infty} x_n \leq 0. $$
My question: Does the same fact hold if we replace sequences with nets? If so, can it be shown via a "diagonal"-type argument?
Yes. A net $(x_\alpha)_{\alpha \in A}$ in a space $X$ is a function $x : A \to X$ defined on a directed set $A$. The collection of all nets is very large - it does not even form a set (but don't let us go here into the fundaments of set theory).
We shall show the following.
Assume that for every $\varepsilon > 0$ there exists a net $(x_\alpha)$ in $X$ such that $$ \limsup_{\alpha} x_\alpha \leq \varepsilon. $$
Then there exists a sequence $(\xi_n)$ in $X$ such that $$ \limsup_{n \to \infty} \xi_n \leq 0. $$
To prove this, we do not need any kind of diagonal argument. In fact, for any $n$ we find a net $(x_\alpha)$ such that $\limsup_{\alpha} x_\alpha \leq 1/n$. In particular, there must exist a $\xi_n = x_{\alpha_n}$ from this net such that $\xi_n \le 2/n$ (otherwise we would have $\limsup_{\alpha} x_\alpha \geq 2/n$). The sequence $(\xi_n) $ clearly satisfies $\limsup_{n \to \infty} \xi_n \leq 0$ because $\xi_n \le 2/k$ for $n \ge k$.