If $(y_\lambda)_{\lambda \in \Lambda}$ is a subnet of $(x_\alpha)_{\alpha \in A}$ and if $A$ is a complete lattice, is $\Lambda$ a complete lattice?

Question

Let $X$ be a set, $(x_\alpha)_{\alpha \in A}$ be a net in $X$ and $(y_\lambda)_{\lambda \in \Lambda}$ be a subnet of $(x_\alpha)_{\alpha \in A}$. If $A$ is a complete lattice (i.e. not only a directed set), then is this true that $\Lambda$ is also a complete lattice ? If not, can we get sufficient conditions on the subnet $(y_\lambda)_{\lambda \in \Lambda}$ ?

Here is my attempt: Let $E \subset \Lambda$. Because $(y_\lambda)_{\lambda \in \Lambda}$ is a subnet of $(x_\alpha)_{\alpha \in A}$, there exists $f: \Lambda \to A$ such that:

1. For every $\lambda \in \Lambda$, $y_\lambda = x_{f(\lambda)}$.
2. For every $\lambda, \lambda' \in \Lambda$, $\lambda \leq_\lambda \lambda'$ implies $f(\lambda) \leq_A f(\lambda')$.
3. For every $\alpha \in A$, there exists $\lambda \in \Lambda$ such that $\alpha \leq_A f(\lambda)$.

Consider $f(E) \subset A$. Since $A$ is a complete lattice, there exists $\bar \alpha \in A$ such that for every $\alpha \in f(E)$, $\alpha \leq_A \bar \alpha$. By the third property, consider $\bar \lambda \in \Lambda$ such that $\bar \alpha \leq_A f(\bar \lambda)$, ... (and here, I do not really know what to do; I have the feeling that it is not going to work).

Thank you for your help !