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.
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.