Solution:
The required maximum is $\frac{1}{2n+2}$. To show that the condition in the statement is not met if $\mu > \frac{1}{2n+2}$, let $U = (0,1) \times (0,1)$, choose a small enough positive $\epsilon$, and consider the configuration $C$ consisting of the $n$ four-element clusters of points $\left(\frac{i}{n+1} \pm \epsilon\right) \times \left(\frac{1}{2} \pm \epsilon\right)$, $i = 1, \ldots, n$, the four possible sign combinations being considered for each $i$. Clearly, every open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, has area at most $\left(\frac{1}{n+1} + \epsilon\right) \cdot \left(\frac{1}{2} + \epsilon\right) < \mu$ if $\epsilon$ is small enough.

We now show that, given a finite configuration $C$ of points in an open unit square $U$, there always exists an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, and has an area greater than or equal to $\mu_0 = \frac{2}{|C|+4}$.

To prove this, usage will be made of the following two lemmas whose proofs are left at the end of the solution.

Lemma 1. Let $k$ be a positive integer, and let $\lambda < \frac{1}{\lfloor k / 2 \rfloor + 1}$ be a positive real number. If $t_{1}, \ldots, t_{k}$ are pairwise distinct points in the open unit interval $(0,1)$, then some $t_{i}$ is isolated from the other $t_{j}$ by an open subinterval of $(0,1)$ whose length is greater than or equal to $\lambda$.

Lemma 2. Given an integer $k \geq 2$ and positive integers $m_{1}, \ldots, m_{k}$,
$$\left\lfloor\frac{m_{1}}{2}\right\rfloor + \sum_{i=1}^{k} \left\lfloor\frac{m_{i}}{2}\right\rfloor + \left\lfloor\frac{m_{k}}{2}\right\rfloor \leq \sum_{i=1}^{k} m_{i} - k + 2$$

Back to the problem, let $U = (0,1) \times (0,1)$, project $C$ orthogonally on the $x$-axis to obtain the points $x_{1} < \cdots < x_{k}$ in the open unit interval $(0,1)$, let $\ell_{i}$ be the vertical through $x_{i}$, and let $m_{i} = |C \cap \ell_{i}|$, $i = 1, \ldots, k$.

Setting $x_{0} = 0$ and $x_{k+1} = 1$, assume that $x_{i+1} - x_{i-1} > (\lfloor m_{i} / 2 \rfloor + 1) \mu_0$ for some index $i$, and apply Lemma 1 to isolate one of the points in $C \cap \ell_{i}$ from the other ones by an open subinterval $x_{i} \times J$ of $x_{i} \times (0,1)$ whose length is greater than or equal to $\mu_0 / (x_{i+1} - x_{i-1})$. Consequently, $\left(x_{i-1}, x_{i+1}\right) \times J$ is an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$ and has an area greater than or equal to $\mu_0$.

Next, we rule out the case $x_{i+1} - x_{i-1} \leq (\lfloor m_{i} / 2 \rfloor + 1) \mu_0$ for all indices $i$. If this were the case, notice that necessarily $k > 1$; also, $x_{1} - x_{0} < x_{2} - x_{0} \leq (\lfloor m_{1} / 2 \rfloor + 1) \mu_0$ and $x_{k+1} - x_{k} < x_{k+1} - x_{k-1} \leq (\lfloor m_{k} / 2 \rfloor + 1) \mu_0$. With reference to Lemma 2, write
$$\begin{aligned} 2 = 2(x_{k+1} - x_{0}) & = (x_{1} - x_{0}) + \sum_{i=1}^{k} (x_{i+1} - x_{i-1}) + (x_{k+1} - x_{k}) \\ & < \left( (\lfloor \frac{m_{1}}{2} \rfloor + 1) + \sum_{i=1}^{k} (\lfloor \frac{m_{i}}{2} \rfloor + 1) + (\lfloor \frac{m_{k}}{2} \rfloor + 1) \right) \cdot \mu_0 \\ & \leq (\sum_{i=1}^{k} m_{i} + 4) \mu_0 = (|C| + 4) \mu_0 = 2 \end{aligned}$$
and thereby reach a contradiction.

Finally, we prove the two lemmas.

Proof of Lemma 1. Suppose, if possible, that no $t_{i}$ is isolated from the other $t_{j}$ by an open subinterval of $(0,1)$ whose length is greater than or equal to $\lambda$. Without loss of generality, we may (and will) assume that $0 = t_{0} < t_{1} < \cdots < t_{k} < t_{k+1} = 1$. Since the open interval $(t_{i-1}, t_{i+1})$ isolates $t_{i}$ from the other $t_{j}$, its length, $t_{i+1} - t_{i-1}$, is less than $\lambda$. Consequently, if $k$ is odd we have $1 = \sum_{i=0}^{(k-1)/2} (t_{2i+2} - t_{2i}) < \lambda (1 + \frac{k-1}{2}) < 1$; if $k$ is even, we have $1 < 1 + t_{k} - t_{k-1} = \sum_{i=0}^{k/2-1} (t_{2i+2} - t_{2i}) + (t_{k+1} - t_{k-1}) < \lambda (1 + \frac{k}{2}) < 1$. A contradiction in either case.

Proof of Lemma 2. Let $I_{0}$, respectively $I_{1}$, be the set of all indices $i$ in the range $2, \ldots, k-1$ such that $m_{i}$ is even, respectively odd. Clearly, $I_{0}$ and $I_{1}$ form a partition of that range. Since $m_{i} \geq 2$ if $i$ is in $I_{0}$, and $m_{i} \geq 1$ if $i$ is in $I_{1}$ (recall that the $m_{i}$ are positive integers),
$$\sum_{i=2}^{k-1} m_{i} = \sum_{i \in I_{0}} m_{i} + \sum_{i \in I_{1}} m_{i} \geq 2|I_{0}| + |I_{1}| = 2(k-2) - |I_{1}|, \quad \text{or} \quad |I_{1}| \geq 2(k-2) - \sum_{i=2}^{k-1} m_{i}$$
Therefore,
$$\begin{aligned} \left\lfloor \frac{m_{1}}{2} \right\rfloor + \sum_{i=1}^{k} \left\lfloor \frac{m_{i}}{2} \right\rfloor + \left\lfloor \frac{m_{k}}{2} \right\rfloor & \leq m_{1} + \left( \sum_{i=2}^{k-1} \frac{m_{i}}{2} - \frac{|I_{1}|}{2} \right) + m_{k} \\ & \leq m_{1} + \left( \frac{1}{2} \sum_{i=2}^{k-1} m_{i} - (k-2) + \frac{1}{2} \sum_{i=2}^{k-1} m_{i} \right) + m_{k} \\ & = \sum_{i=1}^{k} m_{i} - k + 2 \end{aligned}$$

Remark. In case $4n$ is replaced by a positive integer $k$ not divisible by $4$, we do not yet know the maximal $\mu$ satisfying the corresponding condition.