This is a past problem for an Applied Math QR Exam:
Suppose $f_n$, for $n \in \mathbb{N}$, and $f$ are continuous functions on the closed interval $[0,1]$. Assume that $f_n \to f$ as $n \to \infty$ uniformly on the open interval $(0,1)$. Prove that
$$ f_n(0) \to f(0) \quad \text{and} \quad f_n(1) \to f(1). $$
Am I wrong to think the proof is just an $\epsilon/3$ argument, at least for $f_n(0) \to f(0)$? If so, then I think that my proof is lacking subtleties to be precise. In particular, I don't think I can give an exact $N = N(\epsilon)$ in applying the definitions. I'm also not sure that proving $f_n(1) \to f(1)$ is any different, except that maybe in the $\epsilon$-$\delta$ definition for, say, $f$, we must say $1 - x < \delta$ implies $|f(x) - f(1)| < \epsilon$. Lastly, my proof isn't too creative - just application of definitions. So, any more elegant proofs would be of great help, too.
Proof of $f_n(0) \to f(0)$. Our goal is to show that $\forall \epsilon > 0$, $\exists N$ such that $n > N$ implies $|f_n(0) - f(0)| < \epsilon$.
By assumption, we know that $f, f_n \in \mathcal{C}([0,1])$ tells us
$$\forall \epsilon > 0, \exists \delta_0 > 0 \text{ such that } |x| < \delta_0 \Longrightarrow |f(x) - f(0)| < \frac{\epsilon}{3},$$
$$\forall \epsilon > 0, \exists \delta_1 > 0 \text{ such that } |x| < \delta_1 \Longrightarrow |f_n(x) - f_n(0)| < \frac{\epsilon}{3},$$
and that $f_n \to f$ uniformly on $(0,1)$ tells us
$$\forall \epsilon > 0, \exists M \text{ such that } \forall x \in (0,1), n > M \Longrightarrow |f_n(x) - f(x)| < \frac{\epsilon}{3}.$$
Now, for any $x \in (0,1)$, we have
\begin{align*}|f_n(0) - f(0)| &= |f_n(0) - f_n(x) + f_n(x) - f(x) + f(x) - f(0)| \\ &\leq |f_n(x) - f_n(0)| + |f_n(x) - f(x)| + |f(x) - f(0)| \\ &< 3 \cdot \frac{\epsilon}{3} = \epsilon.\end{align*}
So, choosing any $N > M$ and letting $\delta_0, \delta_1 \to 0$ implies $\epsilon \to 0$, and therefore $f_n(0) = f(0)$.