Consider, $u_n(x) = 0$ in $(a, b)$, $u_n(x) = 1, x=a$ or $x = b$. You get a uniform convergence in $(a, b)$ but divergence at the endpoints.
Continuity Edit: Let $s_n$ be the partial sum of $u_i$'s up to term n. Suppose, that the sequence does not converge at $a$, that is, $$\exists\epsilon > 0, \forall n_0 \exists n,m > n_0, |s_n(a) - s_m(a)| > \epsilon \tag{*} \label{*}$$ furthermore, suppose that, $$ \forall n\forall\epsilon'> 0 \ \exists\delta > 0\ | x\in(a, a + \delta) \implies |s_n(x) - s_n(a)| \leq \epsilon' \tag{**} \label{**}$$ We will show that the sequence $s_n$ is not uniformly Cauchy.
Take $\epsilon$ given by $\eqref{*}$. Pick some $n_0$, take the given $n, m$ by $\eqref{*}$. For each of $n, m$ invoke $\eqref{**}$ with $\epsilon' = \epsilon / 4$. Take the minimum of the $\delta$'s given by \eqref{**} and an $x$ in that interval. Then, $$ |s_n(x) - s_m(x)| \geq |s_n(a) - s_m(a)| - |s_n(x) - s_n(a)| - |s_m(x) - s_m(a)| \geq \epsilon / 2$$
Therefore, $$ \exists\epsilon'' > 0, \forall n_0 \exists n,m > n_0, \exists x\in (a,b), |s_n(x) - s_m(x)| > \epsilon''$$ That is, the sequence is not uniformly cauchy.
The rest is showing that a sequence is uniformly cauchy iff it is uniformly convergent. So the statement seems correct.
