The likelihood function is $$L(\theta|\mathbb x)=\begin{cases}\dfrac{1}{\theta ^n},\,\,\,\theta \le x_i \le 2\theta ,\forall i\\0,\,\,\,\,\,\,\,\,\text{otherwise}\end{cases}$$ $$=\begin{cases}\dfrac{1}{\theta ^n},\,\,\,\theta \le x_{(1)} \le x_{(n)} \le2\theta \\0,\,\,\,\,\,\,\,\,\text{otherwise}\end{cases}$$
For $\theta \ge \dfrac {x_{(n)}}{2},L(\theta|\mathbb x)=\dfrac{1}{\theta ^n}$ is a decreasing function in $\theta$. Hence MLE of $\theta$ is $\color{blue}{\hat\theta=\dfrac{X_{(n)}}{2}}$.
Welcome back to MSE.
This is one of those things that once you're explained it correctly the first time, without any gaps in explanation, that it makes sense. Unfortunately, most answers and even professors don't explain all of the details, in my experience.
Suppose $X_1, \dots, X_n$ are independent and distributed $\text{Uniform}(0, \theta)$, with $\theta > 0$.
Let $\mathbf{I}$ denote the indicator function, where
$$\mathbf{I}(\cdot) = \begin{cases}
1, & \cdot \text{ is true} \\
0, & \cdot \text{ is false.}
\end{cases}$$
The probability density function of any of the $X_i$, for $i \in \{1, \dots, n\}$, can be written like so:
$$f_{X_i}(x_i \mid \theta) = \dfrac{1}{\theta}\cdot\mathbf{I}(0<x_i<\theta)\text{.}$$
The likelihood function is thus given by
$$\begin{align}
L(\theta)&=f_{X_1, \dots, X_n}(x_1, \dots, x_n \mid \theta)\\
&=\prod_{i=1}^{n}f_{X_i}(x_i \mid \theta) \\
&= \dfrac{1}{\theta^n}\prod_{i=1}^{n}\mathbf{I}(0 < x_i < \theta)\text{.}
\end{align}$$
The following claim, although used, is often omitted from explanations:
Claim. Let $A$ and $B$ be events. Then $\mathbf{I}(A)\cdot \mathbf{I}(B)=\mathbf{I}(A \cap B)$.
I leave the proof of this to you. Note that $ 0 < x_i < \theta$ is the same as requiring both $x_i > 0$ and $x_i < \theta$. Hence, we write
$$\begin{align}
L(\theta)&=\dfrac{1}{\theta^n}\prod_{i=1}^{n}\mathbf{I}(0 < x_i < \theta) \\
&= \dfrac{1}{\theta^n}\prod_{i=1}^{n}[\mathbf{I}(x_i > 0)\mathbf{I}(x_i < \theta)] \\
&= \dfrac{1}{\theta^n}\prod_{i=1}^{n}[\mathbf{I}(x_i > 0)]\prod_{j=1}^{n}[\mathbf{I}(x_j < \theta)]\text{.}
\end{align}$$
It will be clear why I split the product as above in a bit.
The claim given above is true if we were to extend to an arbitrary number of events as well. Thus,
$$\prod_{i=1}^{n}[\mathbf{I}(x_i > 0)] = \mathbf{I}(x_1 > 0 \cap x_2 > 0 \cap \cdots \cap x_n > 0)$$
and
$$\prod_{j=1}^{n}[\mathbf{I}(x_j < \theta)] = \mathbf{I}(x_1 < \theta \cap x_2 < \theta \cap \cdots \cap x_n < \theta)\text{.}$$
The next claims are often omitted as well from explanations:
Claim 1. Given $x_1, \dots, x_n \in \mathbb{R}$, $x_1, \dots, x_n < k$ if and only if $$x_{(n)}:=\max_{1 \leq i \leq n}x_i < k\text{.}$$
Claim 2. Given $x_1, \dots, x_n \in \mathbb{R}$, $x_1, \dots, x_n > k$ if and only if $$x_{(1)}:=\min_{1 \leq i \leq n}x_i > k\text{.}$$
Thus
$$\prod_{i=1}^{n}[\mathbf{I}(x_i > 0)] = \mathbf{I}(x_1 > 0 \cap x_2 > 0 \cap \cdots \cap x_n > 0) = \mathbf{I}(x_{(1)} > 0)$$
and
$$\prod_{j=1}^{n}[\mathbf{I}(x_j < \theta)] = \mathbf{I}(x_1 < \theta \cap x_2 < \theta \cap \cdots \cap x_n < \theta) = \mathbf{I}(x_{(n)} < \theta)\text{.}$$
The likelihood function is thus
$$L(\theta) = \dfrac{1}{\theta^n}\mathbf{I}(x_{(1)} > 0)\mathbf{I}(x_{(n)} < \theta)\text{.}\tag{*}$$
Now, consider the above as a function of $\theta$. For all intents and purposes, $\mathbf{I}(x_{(1)} > 0)$ is irrelevant when it comes to maximization of $L$ with respect to $\theta$, because it is independent of $\theta$. So, the part that really matters is
$$L(\theta) \propto \dfrac{1}{\theta^n}\mathbf{I}(x_{(n)} < \theta) = \dfrac{1}{\theta^n}\mathbf{I}(\theta > x_{(n)})\text{.}\tag{**}$$
Generally, when doing maximum-likelihood estimation, we assume that the observed $x_i$ fall within the support of the given distribution, so we'll just assume $x_{(1)} > 0$.
Remember to view (**) as a function of $\theta$. If $\theta \leq x_{(n)}$, note that $L(\theta) = 0$ because of the indicator function. This is not the maximized value of $L$; $L$ is, at its crux, a probability density function: $0$ is in fact the smallest value that a probability density function can take.
So, in attempting to maximize $L$, suppose that $\theta > x_{(n)}$. For $n$ fixed, we obtain
$$L(\theta) \propto\dfrac{1}{\theta^n}\text{.}$$
Now, note that $\dfrac{1}{\theta^n}$ is indeed a decreasing function of $\theta$ with $n$ fixed. Thus, we must make $\theta$ as small as possible, given our restriction of $\theta > x_{(n)}$.
Note. Technically, no such $\theta$ exists (because $\theta$ is strictly greater than $x_{(n)}$ per our assumptions). This is often ignored in many textbooks.
Most textbooks will then say that the maximum likelihood estimator of $\theta$ is
$$\hat{\theta}_{\text{MLE}} = X_{(n)}\text{.}$$
Note. Technically, the above result is false. The MLE does not exist, because $\theta$ cannot take on the value $x_{(n)}$ itself. For this answer to be correct, the support of the uniform PDF must include $\theta$ itself (because the maximum likelihood estimator equals one of the $X_i$). The reason for this is discussed in the Lecture 2: Maximum Likelihood Estimators from MIT OpenCourseWare 18-443 Statistics for Applications, found here. As the question currently stands, $(0, \theta)$ should be $(0, \theta]$.
Best Answer
I think the problem assumes $c$ and $d$ are known constants, and $\theta$ is the only unknown.
The likelihood is $(d-c)^{-n} \theta^{-n} \prod_{i=1}^n 1_{\{\theta c \le x_i \le \theta d\}}$.
Note that each indicator can be rewritten as $1_{\{x_i/d\le \theta \le x_i/c\}}$, so we need to maximize the likelihood over $\theta \in [x_{(n)}/d, x_{(1)}/c]$. Since $\theta^{-n}$ is decreasing in $\theta$, the MLE is $x_{(n)}/d$. The result is intuitive: the smaller the $\theta$, the smaller the interval $[\theta c, \theta d]$, so the density is larger on this interval. The MLE chooses the smallest $\theta$ such that all the points $x_i$ lie in the interval.