A sequence $f_n to f$ uniformly on $(0,1)$ where $f, f_n in C([0,1])$ implies $f_n(0) to f(0)$, $f_n(1) to f(1)$.

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 $f_n(0) = f(0)$.