The answer to this question is negative.

As a counterexample, consider the one-point extension $P:=2^{<\omega}\cup\{\infty\}$ of the binary tree $2^{<\omega}=\bigcup_{n\in\omega}2^n$. Here $2$ is the ordinal $\{0,1\}$ and $2^n$ is the set of binary sequences of length $n$.

The partial order on $P$ is defined by two conditions:

(i) $p\le\infty$ for all $p\in P$ and

(ii) $x\le y$ for $x,y\in 2^{<\omega}$ if the sequence $y$ is an extension of the sequence $x$, that is $x=y|n$ for $n=|x|$ where $|x|$ is the unique number such that $x\in 2^{|x|}\subset 2^{<\omega}$.

Now we describe the topology $\tau_o(P)$.

Claim 1. If a set-filter $\mathcal F$ on $P$ converges to a point $p\in 2^{<\omega}$, then $\{p\}\in\mathcal F$.

Proof. It follows from $\mathcal F\to p$ that $\sup\mathcal F^l=p=\inf \mathcal F^u$. The equality $p=\sup\mathcal F^l$ implies that $\mathcal F^l\subset{\downarrow}p$. Taking into account that ${\downarrow}p$ is a finite linearly ordered set, we conclude that $p\in\mathcal F^l$ and hence $p\in F^l$ for some $F\in\mathcal F$. Then $F\subset{\uparrow}p$ and hence ${\uparrow}p\in\mathcal F$.

The equality $p=\inf\mathcal F^u$ implies that $\mathcal F^u\subset{\uparrow}p$. Taking into account that $P$ contains no strictly decreasing sequences, we conclude that $p=\inf A$ for some finite subset $A\subset\mathcal F^u$. Since $\mathcal F$ is closed under taking finite intersections, $A\subset F^u$ for some $F\in\mathcal F$ and then $p=\inf A\in F^u$, which implies $F\subset {\downarrow}p$ and ${\downarrow}p\in \mathcal F$.

Finally, we see that $\{p\}={\uparrow}p\cap{\downarrow}p\in\mathcal F$.

Corollary 1. Each point $p\in 2^{<\omega}$ is isolated in the topology $\tau_o(P)$.

Claim 2. If a set-filter $\mathcal F$ on $P$ converges to $\infty$, then either $\{\infty\}\in\mathcal F$ or $\{\sup F^l:F\in\mathcal F\}$ is an infinite chain in $2^{<\omega}$.

Proof. Assume that $\{\infty\}\notin\mathcal F$. First we show that for any $F\in\mathcal F$ the set $F^l$ is linearly ordered. Indeed, assuming that $F^l$ contains two incomparable elements $x,y$, we conclude that $$\emptyset\ne F\in{\uparrow}x\cap{\uparrow}y=\{\infty\}$$ and hence $\{\infty\}=F\in\mathcal F$, which contradicts our assumption. So, the set $F^l$ is linearly ordered and hence has $\sup F^l\in F^l\subset 2^{<\omega}$.

To see that for any sets $F,E\in\mathcal F$ the elements $\sup F^l$ and $\sup E^l$ are comparable, observe that they belong to the linearly ordered set $(F\cap E)^l$.

Therefore, the set $L=\{\sup F^l:F\in\mathcal F\}$ is linearly ordered. Assuming that this set is finite, we can conclude that $\infty=\sup\mathcal F^l=\sup L\in L$, which contradicts the inclusion $L\subset 2^{<\omega}$.

Corollary 2. If an open set $U\in\tau_0(P)$ contains $\infty$, then $2^{<\omega}\setminus U$ is finite.

Proof. To derive a contradiction, assume that the complement $C:=2^{<\omega}\setminus U$ is ifinite. Applying the Konig Tree Lemma, we can find an infinite linearly ordered set $L\subset C$. Consider the set-filter $$\mathcal F=\{F\subset P:L\setminus F\mbox{ is finite}\}$$ and observe that $\mathcal F\to\infty$ and $U\notin \mathcal F$, which contradicts the definition of the topology $\tau_0(P)$.

Claim 3. A subset $U\subset P$ containing $\infty$ is $\tau_0(P)$-open if and only if the complement $P\setminus U$ is finite.

Proof. The "only if" part is proved in Corollary 2. To prove the "if" part, assume that $P\setminus U$ is finite. The $\tau_o(P)$-openness of $U$ will follow from Claims 1 as soon as we show that for any filter $\mathcal F\to\infty$ the set $U$ belongs to $\mathcal F$. By Claim 2, either $\{\infty\}\in\mathcal F$ or else $\{\sup F^l:F\in\mathcal F\}$ is an infinite chain in $2^{<\omega}$. In the first case the inclusions $U\supset\{\infty\}\in\mathcal F$ imply that $U\in\mathcal F$. In the second case, we can observe that the set $D:=\bigcup_{x\in 2^{<\omega}\setminus U}{\downarrow}x$ is finite, so we can find a set $F\in\mathcal F$ such that $\sup F^l\notin D$ and conclude that $F\cap D=\emptyset$ and $U\supset P\setminus D\supset F\in\mathcal F$ implies that $U\in\mathcal F$.

Now we are able to prove that the poset $P$ provides a counterexample to the original question.

By Claim 3, for every $n\in\mathbb N$ the set $U_n:=P\setminus\bigcup_{k\le n}2^k$ belongs to the topology $\tau_0(P)$. Then the set-filter $\mathcal F=\{F\subset P:\exists n\in\mathbb N\;U_n\subset F\}$ topologically converges to $\infty$. On the other hand, the set $\{\sup F^l:F\in\mathcal F\}=\{\sup U_n^\ell:n\in\mathbb N\}$ coincides with the singleton $2^0$ and hence $\mathcal F\not\to\infty$ by Claim 2.