The answer to this question is affirmative:
Theorem. There exists a countable set $X$ and an uncountable family $\mathcal F$ of self-functions of $X$ such that the poset $T_2(\mathcal F)$ has no minimal elements.
Here $T_2(\mathcal F)$ is the poset of all Hausdorff topologies on $X$ making all functions $f\in\mathcal F$ continuous.
Proof. To construct the space $X$ and the family $\mathcal F$, take any Hausdorff $(\omega_1,\omega_1)$-gap on $\omega$, which is a pair $\big((A_\alpha)_{\alpha\in\omega_1},(B_\alpha)_{\alpha\in\omega_1}\big)$ of families of infinite subsets of $\omega$ satisfying the following two conditions:
(H1) for any $\alpha<\beta<\omega_1$ we have $A_\alpha\subset^* A_\beta\subset^* B_\beta\subset^* B_\alpha$;
(H2) for any set $C\subset\omega$ one of the sets $\{\alpha\in \omega_1:A_\alpha\subset^* C\}$ or $\{\alpha\in\omega_1:C\subset^* B_\alpha\}$ is at most countable.
Here the notation $A\subset^* B$ means that the complement $A\setminus B$ is finite.
It is well-known that Hausdorff $(\omega_1,\omega_1)$-gaps do exist in ZFC.
Let $X=\{-\infty,+\infty\}\cup\omega$ and for every $\alpha\in\omega_1$ consider the functions $f_\alpha,g_\alpha:X\to X$ defined by
$$f_\alpha(x)=\begin{cases}
x&\mbox{if $x\in \{-\infty\}\cup B_\alpha$};\\
+\infty&\mbox{otherwise};
\end{cases}
$$
and
$$g_\alpha(x)=\begin{cases}
x&\mbox{if $x\in \{+\infty\}\cup (\omega\setminus A_\alpha)$};\\
-\infty&\mbox{otherwise}.
\end{cases}
$$
For every $a,b\in\omega$ let $p_{a,b}:X\to X$ be the function defined by
$$p_{a,b}(x)=\begin{cases}
b&\mbox{ if $x=a$},\\
a&\mbox{ if $x=b$},\\
x&\mbox{otherwise}.
\end{cases}
$$
We claim that the family of functions $\mathcal F=\{f_\alpha\}\cup\{g_\alpha\}_{\alpha\in\omega_1}\cup\{p_{a,b}\}_{a,b\in\omega}$ has the required property: the poset $T_2(\mathcal F)$ has no minimal elements.
Indeed, take any Hausdorff topology $\sigma$ on $X$ in which every map $f\in\mathcal F$ is continuous. The continuity of the maps $p_{a,b}$, $a,b\in\omega$, implies that each point of the set $\omega$ is isolated in the topological space $(X,\sigma)$.
We claim that for every $\alpha\in\omega_1$ the set $W_\alpha^-:=B_\alpha\cup\{-\infty\}$ is a neighborhood of $-\infty$ and $W_\alpha^+:=(X\setminus A_\alpha)\cup\{+\infty\}$ is a neighborhood of $+\infty$ in the topology $\sigma$.
For this choose any two disjoint neighborhoods $U_-,U_+\in\sigma$ of $-\infty$ and $+\infty$, respectively. By the continuity of the maps $f_\alpha$ and $g_\alpha$, there are neighborhoods $V_-,V_+\in\sigma$ of $-\infty$ and $+\infty$ such that $f_\alpha(V_-)\subset U_-$ and $g_\alpha(V_+)\subset U_+$. Then $$V_-\subset f_\alpha^{-1}(U_-)\subset f_\alpha^{-1}(X\setminus\{+\infty\})\subset \{-\infty\}\cup B_\alpha$$
and $$V_+\subset g_\alpha^{-1}(U_+)\subset g_\alpha^{-1}(X\setminus\{-\infty\})\subset\{+\infty\}\cup(\omega\setminus A_\alpha).$$
Therefore, the topology $\sigma$ contains the topology $\tau$ defined in the answer to this MO problem. Repeating the argument from this answer it can be shown that $\sigma$ is not a minimal element of the poset $T_2(\mathcal F)$.
However, for the interested reader let us write more details.
By the condition (H2), one of the sets $A=\{\alpha\in\omega_1:A_\alpha\subset^* U_-\}$ or $B=\{\alpha\in\omega_1:U_-\setminus\{-\infty\}\subset^* B_\alpha\}$ is countable.
First we assume that the set $A$ is countable. In this case we can find a countable ordinal $\alpha$ such that $A_\alpha\not\subset^*U_-$.
Consider the topology $\sigma'$ on $X$ consisting of sets $W\subset X$ satisfying two conditions:
$\bullet$ if $-\infty\in W$, then there exists $U\in\sigma$ such that $-\infty\in U\cup A_\alpha\subset^* W$;
$\bullet$ if $+\infty\in W$, then there exists $U\in\sigma$ such that $+\infty\in U\subset^* W$.
It is clear that $\sigma'\subset\sigma$ and $\sigma'\ne\sigma$ (because $U_-\in\sigma\setminus\sigma'$).
The topology $\sigma'$ is Hausdorff since $U_-\cup A_\alpha$ and $U_+\setminus A_\alpha=U_+\cap W^+_\alpha$ are disjoint neighborhoods of $-\infty$ and $+\infty$, respectively.
To see that $\sigma'\in T_2(\mathcal F)$, observe that for any $a,b\in \omega$ the permutation $p_{a,b}$ is continuous as the points $a,b$ are isolated in the topology $\sigma'$. Next, for every ordinal $\beta\in\omega_1$ the inclusion $A_\alpha\subset^* B_\beta$ implies that the map $f_\beta$ is continuous at $-\infty$. The continuity of the map $f_\beta$ at $+\infty$ in the topology $\sigma$ implies the continuity of this map at $+\infty$ in the topology $\sigma'$.
The continuity of the map $g_\beta$ at $-\infty$ in the topology $\sigma'$ follows from the inclusion $g_\beta(x)\in\{-\infty,x\}$ holding for all $x\in X$.
The continuity of $g_\beta$ at $+\infty$ in the topology $\sigma'$ follows from the continuity of $g_\beta$ at $+\infty$ in the topology $\sigma$ and the definition of the topology $\sigma'$.
Now consider the case of countable set $B:=\{\alpha\in\omega_1:U_-\setminus\{-\infty\}\subset^* B_\alpha\}$. In this case we can find a countable ordinal $\alpha$ such that $U_-\setminus\{-\infty\}\not\subset^* B_\alpha$ and hence $\omega\setminus B_\alpha\not\subset^* U_+$.
In this case we can consider the topology $\sigma'$ on $X$ consisting of sets $W\subset X$ satisfying two conditions:
$\bullet$ if $-\infty\in W$, then there exists $U\in\sigma$ such that $-\infty\in U\subset^* W$;
$\bullet$ if $+\infty\in W$, then there exists $U\in\sigma$ such that $+\infty\in U\cup(\omega\setminus B_\alpha)\subset^* W$.
It is clear that $\sigma'\subset\sigma$ and $\sigma'\ne\sigma$ (because $U_+\in\sigma\setminus\sigma'$).
The topology $\sigma'$ is Hausdorff since $U_-\cap B_\alpha$ and $U_+\cup(\omega\setminus B_\alpha)$ are disjoint neighborhoods of $-\infty$ and $+\infty$, respectively. By analogy with the case of countable set $A$, we can show that each map $f\in\mathcal F$ remains continuous in the topology $\sigma'$.
In both cases we have constructed a strictly weaker topology $\sigma'\subset \sigma$ in $T_2(\mathcal F)$ witnessing that the topology $\sigma$ is not minimal in the poset $T_2(\mathcal F)$.
