EDIT: If you want discontinuous functions whose sum converges uniformly to a discontinuous function, consider the characteristic function of the rationals:
$$
\lambda(x) = \left\{
\begin{array}{lr}
1 & : x \in \mathbb{Q}\\
0 & : x \notin \mathbb{Q}
\end{array}
\right.$$
Then let $\lambda_n = \frac{1}{2^{n+1}} \cdot \lambda$. The sum will indeed converge uniformly to $\lambda$ but will clearly be discontinuous. In fact if $\lambda$ is any discontinuous function, this construction will give you your desired result.
However, if the $u_n$'s are all continuous, this cannot happen. Let
$$g_n = \sum_{k=1}^{n} u_k$$
Then each $g_i$ is countinuous being the sum of continuous functions. But since $g_n \rightarrow u$ uniformly, it must be the case that $u$ is continuous as well. Here's a quick proof:
Suppose we want to show that $u$ is continuous at some $x_0 \in I$. Fix some $\epsilon > 0$. Find $N$ large enough such that $|g_n (x) - u(x) | < \frac{\epsilon}{3}$ for all $x \in I$ and $n \ge N$. Choose some $\delta$ such that $| g_N (x) - g_N (x_0) | < \frac{\epsilon}{3}$ whenever $|x - x_0 | < \delta$, which we can do as $g_N$ is continuous.
Now we have that:
$$|u(x) - u(x_0)| \le |u(x) - g_N(x)| + |g_N (x) - g_N (x_0) | + |g_N(x_0) - u(x_0)| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon $$
by applying the triangular inequality. Hence $u$ is continuous at $x_0$.