Let $f_n:[a, b] \rightarrow \mathbb{R}$ be a sequence of continuous functions which is uniformly bounded i.e. $||f_n||_{L^{\infty}} \leq M <\infty$ and satisfies $f_n(a)=A$ for all $n\in \mathbb{N}.$ In addition for every $\epsilon >0,$ the function $f_n$ also satisfies the following equicontinuity estimate, $$|f_n(x)-f_n(y)| \leq c(\epsilon)|x-y| \quad \quad \quad \text{ for all } x,y\in [a+\epsilon, b],$$ with $c(\epsilon) \rightarrow \infty$ as $\epsilon \rightarrow 0^+.$
Now can we conclude that up to a subsequence $f_n \rightarrow f$ pointwise where $f \in C([0,b])$ and $f(a)=A?$
P.S.: If the answer is yes, a clean proof of the same would be appreciated.
You have convergence at $a$ for free, you have uniform convergence up to a subsequence on $[c,b]$ for any $a<c \leq b$ from Arzela-Ascoli. By a diagonal argument (extract a subsequence on, say, $[a+(b-a)/(n+1),b]$ and extract the diagonal of the resulting array), you have pointwise convergence up to a subsequence on $[a,b]$. The pointwise limit is guaranteed to be continuous everywhere except $a$.
To finish up, you can't be sure about $a$ because you don't have the flexibility to invoke Arzela-Ascoli on the full $[a,b]$ with these hypotheses. So you go hunting for counterexamples to see whether the result might just be negative, and you arrive at $[a,b]=[0,1],f_n(x)=(1-x)^n$ as a case where $f$ is not continuous at $a$.