Assume $(x_t)_{t\in\Lambda}$ is a bounded net in $\mathbb{C}$ that doesn't convergent to $0$. Is there a subnet of it and an $\epsilon$ such that is outside of $B(0,\epsilon)=\{ x\in \mathbb{C}: \lVert x\rVert<\epsilon\} $?
This is true for sequences but for net I can't find a counterexample.
Since $x_t$ doesn't converge to $0$, there is $\epsilon>0$ such that for all $s\in\Lambda$ there is $t\in\Lambda$ such that $t\geq s$ and $|x_{t}|>\epsilon$.
Let's use this to choose the subnet in an orderly manner.
Let $\preceq$ be a well order of $\Lambda$ and $o$ the first element in this order. Choose $t_o\geq o$ with $|x_{t_0}|>\epsilon$.
Assume that we have defined $t_j$ for all $j\preceq i$ with $j\neq i$. If there is already some $t_j\geq i$ defined for some $j\preceq i$, either skip $i$ or define $t_i$ to be one such $t_j$. By transfinite induction we have defined a subset $\Gamma$ of $\Lambda$ of all values of $t_i$ chosen. This set ordered with the original order $\leq$ of $\Lambda$ is directed. In fact, if $t_i,t_j\in\Gamma$ then there are some $i',j'\in\Lambda$ such that $t_i\leq i'$ and $t_j\leq j'$. Since $(\Lambda,\leq)$ is directed, then there is $k\in\Lambda$ with $k\geq i'$ and $k\geq j'$. But then there is $t_k\in\Gamma$ with $t_k\geq k$. This implies that $t_k\geq t_i$ and $t_k\geq t_j$ by transitivity. Therefore $(\Gamma,\leq)$ is also directed.
By construction, for all elements $r\in\Gamma$ we have $|x_r|>\epsilon$.
The restriction of $x$ to $\Gamma$ is a subnet. In fact, if $i\in\Lambda$ then $\{k\in\Gamma:\ k\geq t_i\}\subset\{j\in\Lambda:\ j\geq i\}$, since $k\geq t_i\geq i$ by construction, and this implies $k\geq i$ by transitivity.