I tried finding a proof without considering characteristic functions. Excess kurtosis does the trick. Here's the two-line answer: $\text{Kurt}(U) = \text{Kurt}(X + Y) = \text{Kurt}(X) / 2$ since $X$ and $Y$ are iid. Then $\text{Kurt}(U) = -1.2$ implies $\text{Kurt}(X) = -2.4$ which is a contradiction as $\text{Kurt}(X) \geq -2$ for any random variable.
Rather more interesting is the line of reasoning that got me to that point. $X$ (and $Y$) must be bounded between 0 and 0.5 - that much is obvious, but helpfully means that its moments and central moments exist. Let's start by considering the mean and variance: $\mathbb{E}(U)=0.5$ and $\text{Var}(U)=\frac{1}{12}$. If $X$ and $Y$ are identically distributed then we have:
$$\mathbb{E}(X + Y) = \mathbb{E}(X) + \mathbb{E}(Y) = 2 \mathbb{E}(X)= 0.5$$
So $\mathbb{E}(X) = 0.25$. For the variance we additionally need to use independence to apply:
$$\text{Var}(X+Y) = \text{Var}(X) + \text{Var}(Y) = 2 \text{Var}(X) = \frac{1}{12}$$
Hence $\text{Var}(X) = \frac{1}{24}$ and $\sigma_X = \frac{1}{2\sqrt{6}} \approx 0.204$. Wow! That is a lot of variation for a random variable whose support ranges from 0 to 0.5. But we should have expected that, since the standard deviation isn't going to scale in the same way that the mean did.
Now, what's the largest standard deviation that a random variable can have if the smallest value it can take is 0, the largest value it can take is 0.5, and the mean is 0.25? Collecting all the probability at two point masses on the extremes, 0.25 away from the mean, would clearly give a standard deviation of 0.25. So our $\sigma_X$ is large but not impossible. (I hoped to show that this implied too much probability lay in the tails for $X + Y$ to be uniform, but I couldn't get anywhere with that on the back of an envelope.)
Second moment considerations almost put an impossible constraint on $X$ so let's consider higher moments. What about Pearson's moment coefficient of skewness, $\gamma_1 = \frac{\mathbb{E}(X - \mu_X)^3}{\sigma_X^3} = \frac{\kappa_3}{\kappa_2^{3/2}}$? This exists since the central moments exist and $\sigma_X \neq 0$. It is helpful to know some properties of the cumulants, in particular applying independence and then identical distribution gives:
$$\kappa_i(U) = \kappa_i(X + Y) = \kappa_i(X) + \kappa_i(Y) = 2\kappa_i(X)$$
This additivity property is precisely the generalisation of how we dealt with the mean and variance above - indeed, the first and second cumulants are just $\kappa_1 = \mu$ and $\kappa_2 = \sigma^2$.
Then $\kappa_3(U) = 2\kappa_3(X)$ and $\big(\kappa_2(U)\big)^{3/2} = \big(2\kappa_2(X)\big)^{3/2} = 2^{3/2} \big(\kappa_2(X)\big)^{3/2}$. The fraction for $\gamma_1$ cancels to yield $\text{Skew}(U) = \text{Skew}(X + Y) = \text{Skew}(X) / \sqrt{2}$. Since the uniform distribution has zero skewness, so does $X$, but I can't see how a contradiction arises from this restriction.
So instead, let's try the excess kurtosis, $\gamma_2 = \frac{\kappa_4}{\kappa_2^2} = \frac{\mathbb{E}(X - \mu_X)^4}{\sigma_X^4} - 3$. By a similar argument (this question is self-study, so try it!), we can show this exists and obeys:
$$\text{Kurt}(U) = \text{Kurt}(X + Y) = \text{Kurt}(X) / 2$$
The uniform distribution has excess kurtosis $-1.2$ so we require $X$ to have excess kurtosis $-2.4$. But the smallest possible excess kurtosis is $-2$, which is achieved by the $\text{Binomial}(1, \frac{1}{2})$ Bernoulli distribution.