Using Central Limit Theorem to approximate.

While $X$ has a uniform distribution, $S$ does not. Therefore you cannot use directly the mean and variance formulas that apply for a uniform distribution to find $E(S)$ and $V(S),$ as you have tried to do.

Here is an outline of what you need to do.

By the formulas for a uniform distribution, you have $E(X_i) = 1/2$ and $V(X_i) = 1/12.$ To find $E(S)$ you proceed as follows. using a rule that says "expectation of sum is sum of expectations":

$$E(S_{100}) = E(X_1 + X_2 + \cdots + X_{100}) = \sum_{i=1}^{100} E(X_i) = 100(1/2) = 50.$$

Because the $X_i$ are independent, you can use a similar method to add the 100 variances to get $V(S_{100}) = 100(1/12) = 100/12 = 8.3333.$ Then $SD(S_{100}) = \sqrt{100/12} = 2.887.$

Now, by the Central Limit Theorem, $S_{100} \approx W,$ where $W \sim \mathrm{Norm}(\mu = 50, \sigma = 2.887).$ So your problem becomes the evaluation of $P(40 \le W \le 60) \approx P(40 \le S_{100} \le 60).$ I assume you know how to standardize $W$ and use standard normal tables to evaluate $P(40 \le W \le 60)$.

Using software I get an answer that is very nearly 1. This makes sense because a normal distribution has almost all of its area within three standard deviations of the mean. This would be the interval $50 \pm 3(2.887)$ or roughly $(41.3, 58.7),$ which is contained in $(40, 60).$