The p.d.f for one $x_i$ is given as
$$
f(x| \theta) = \begin{cases}
\frac{1}{\theta} & & \text{if } 0 \leq x \leq \theta \\
0 & & \text{otherwise}
\end{cases}
$$
Let's call $\vec{x} = (x_1, ..., x_n)$.
The $n$ observations are i.i.d. so the likelihood of observing the $n$-vector $\vec{x} = (x_1, ... x_n)$ is the product of the component-wise probabilities. Ignoring the issue of support for the moment, note that this product can be simply written as a power:
$$
f(\vec{x}| \theta) = \prod_i^n \frac{1}{\theta} = \frac{1}{\theta^n} = \theta^{-n}
$$
Next, we turn our attention to the support of this function. If any single component is outside its interval of support $(0, 1/\theta)$, then its contribution to this equation is a 0 factor, so the product of the whole will be zero. Therefore $f(\vec{x})$ only has support when all components are inside $(0, 1/\theta)$.
$$
f(\vec{x}| \theta) = \begin{cases}
\theta^{-n} & & \text{if } \forall i, \ 0 \leq x_i \leq \theta \\
0 & & \text{otherwise}
\end{cases}
$$
By definition, this is also our likelihood:
$$
\mathcal{L}(\theta; \vec{x}) = f(\vec{x}| \theta) = \begin{cases}
\theta^{-n} & & \text{if } \forall i, \ 0 \leq x_i \leq \theta \\
0 & & \text{otherwise}
\end{cases}
$$
The MLE problem is to maximize $\mathcal{L}$ with respect to $\theta$. But because $\theta > 0$ (given in the title of the problem) then $\theta^{-n} > 0$ therefore 0 will never be the maximum. Thus, this is a constrained optimization problem:
$$
\hat{\theta} = \text{argmin}_\theta \,\, \theta^{-n} \text{ s.t. } \forall i \,\, 0 \leq x_i \leq \theta
$$
This is easy to solve as a special case so we don't need to talk about the simplex method but can present a more elementary argument. Let $t = \text{max} \,\, \{x_1,...,x_n\}$. Suppose we have a candidate solution $\theta_1 = t - \epsilon$. Then let $\theta_2 = t - \epsilon/2$. Clearly both $\theta_1$ and $\theta_2$ are on the interior of the feasible region. Furthermore we have $\theta_2 > \theta_1 \implies \theta_2^{-n} < \theta_2^{-n}$. Therefore $\theta_1$ is not at the minimum. We conclude that the minimum cannot be at any interior point and in particular must not be strictly less than $t$. Yet $t$ itself is in the feasible region, so it must be the minimum. Therefore,
$$\hat{\theta} = \text{max} \,\, \{x_1,..., x_n\}$$
is the maximum likelihood estimator.
Note that if any observed $x_i$ is less than 0, then $\mathcal{L}$ is a constant 0 and the optimization problem has no unique solution.