Let $\tau$ be the standard topology on $\mathbb R$ and let $U=(a,b)$ a nonempty open interval.
There are two notions of maximal:
Answer 1: If Maximal sub-topology means: finer than all other sub-topologies stable under $U$.
We'll construct two topologies $\tau_a$ and $\tau_b$ such that $U$ is not an open set in either, but such that the join $\tau_a \vee \tau_b$ is the usual topology $\tau$ on $\mathbb R$. This would answer your question in the negative.
Choose $p,r$ distinct points in $(a,b)$. The idea is to have $\tau_a$ the simplest topology where $p$ is topologically indistinguishable from $a$, and similarly $\tau_b$ the simplest topology in which $r$ is indistinguishable from $b$. So, for example, taking the kolmogorov quotient of $\tau_a$, glues $p$ and $a$.
Here's an explicit description of the neighborhood system of $\tau_a$: Every $x\in \mathbb R\backslash\{p,a\}$ has the same neighborhoods as in $\tau$, and for $y\in\{p,a\}$, the neighborhoods of $y$ are the sets $N_a\cup N_p$ where $N_a$ is a neighborhood of $a$ and $N_p$ a neighborhood of $p$ in $\tau$.
$U\not\in\tau_a$ because $U$ is not in the neighborhood of $p$. Indeed, any open set containing $p$ must also contains $a$.
A similar construction for $\tau_b$ and $r$.
It's not hard to see that $\tau_a\vee\tau_b=\tau$. Indeed, $\tau=\{A\cap B,\ A\in\tau_a,\ B\in\tau_b\}$.
Answer 2: If Maximal sub topology means: is contained in no other sub topology stable under $U$.
This time the answer is positive. The idea is to construct the simplest topology $\tau'_a$ such that $p\in U$ is 'asymmetrically glued' to $a$.
An explicit description of the neighborhood system of $\tau'$ is as follows: For $x\in\mathbb R\backslash\{p\}$, the neighborhoods of $x$ are the same as in $\tau$. The neighborhoods of $p$ are the sets $N_a\cup N_p$ where $N_a$ is a neighborhood of $a$ and $N_p$ a neighborhood of $p$ in $\tau$.
Again $U\not\in\tau'$ because it's not a neighborhood of $p$. Any open set containing $p$ must also contain $a$.
On the other hand, extending $\tau'$ by an open set $V\in\tau\backslash \tau'$ retrieves $\tau$. Indeed there's an open interval $I\subset V$ such that $p\in I$ and $a\not\in I$. And any open sub-interval of $I$ is of the form $V\cap A$ with $A\in\tau'$. So we get back all neighborhoods of $p$.
Remark: In the first answer, two points $p,r\in U$ are used to keep things nice and symmetric. But, following the second construction, one point is in fact sufficient. Also, $a$ and $b$ could be replaced with any points not in $U$.
