Let $(a_d)_{d\in D}$ be a real net such that $\lim_{d\in D}a_d=∞$. Is there a cofinal set $D'\subset D$ such that $(a_d)_{d\in D'}$ is increasing? Let $(D, \geq)$ be a directed set, and let
$(a_d)_{d\in D}$ be a real-valued net satisfying
$$
\lim_{d\in D}a_d =+\infty.
$$
Can we find a cofinal subset $D'\subset D$ such that the restricted net $(a_d)_{d\in D'}$ is increasing, i.e., $a_{t}\geq a_d$ whenever $t,d\in D'$ and $t\geq d$?
The result is easily proved under the additional assumption that there exists a cofinal sequence $(t_n)_{n\in \mathbb N}$ in $D$, but I would like to avoid this assumption.
 A: This is a partial answer offered in the hope that it  might  help those searching for a counter example:

The claim is that, if $D$ contains a linearly ordered,  cofinal  subset, then the answer is positive.
In order to prove it we may of course assume that $D$ itself is linearly ordered.
It is then easy to recursevely   build an increasing sequence
$$
  i_0<i_1<\cdots <i_n<\cdots
  $$
in $D$ such that, for each $k\in {\mathbb N}\setminus\{0\}$, one has that
$$
  i\geq i_k \Rightarrow a_i> \max\{k, a_{i_{k-1}}\}.
  \tag 1
  $$
Letting $D'=\{i_0, i_1,  i_2, \ldots \}$, it is obvious that the restriction of the net to $D'$ is (strictly) increasing, so it suffices to
prove that $D'$ is cofinal.
For this, pick any $i$ in $D$, and let $k$ be any integer such that $k> a_i$.
Observe that one cannot have that $i\geq i_k$  or, otherwise,  (1) would imply that $a_i>k$.  The assumption that $D$ is
linearly ordered then gives $i<i_k$, hence proving $D'$ to be cofinal.

In conclusion,  should one be looking for a counter-example,  it will be necessary to consider directed sets $D$ not
possessing any linearly ordered,  cofinal subset.
One such example is the directed set
$\mathscr P_{\text{fin}}({\mathbb R})$
consisting of all finite subsets of ${\mathbb R}$,  under the
order of inclusion.   If $\mathscr L$ is a linearly ordered subset of
$\mathscr P_{\text{fin}}({\mathbb R})$,  one has that
$$
  \bigcup_{X\in \mathscr L}X
  $$
is countable,  so $\mathscr L$ cannot be cofinal.
