Uniform convergence and differentiability, question about the proof

I have a question to the following proof of the well-known theorem.

Theorem. Let $\{f_n \}$ be a sequence of functions converging to $f$ pointwise on $[a,b]$. If each $f_n$ is differentiable with ${f_n}^{\prime} \to g$ uniformly for some function $g$, then $f^{\prime} = g.$

Proof. Let $\epsilon > 0$ be given. Take any $c \in (a,b)$. Since ${f_n}^{\prime} \to g$,$$ \exists N_1 \in \mathbb{N}:\forall m \in \mathbb{N} \; \; m \ge N_1 \implies \left| {f_m}^{\prime}(c) - g(c) \right| < \frac{\epsilon}{3}$$

Since every convergent sequence is a Cauchy sequence,$$\exists N_2 \in \mathbb{N}: \forall n,m \in \mathbb{N} \; \; n,m \ge N_2 \implies \left| {f_n}^{\prime}(x) - {f_m}^{\prime}(x) \right| < \frac{\epsilon}{3} $$ for any $x \in (a,b)$. Take $N := \text{max}(N_1, N_2)$. Since $f_N$ is differentiable,$$\exists \delta > 0: \forall x \in [a,b] \; \; \left| x - c \right| < \delta \implies \left| \frac{f_N(x) - f_N(c)}{x-c} - {f_N}^{\prime}(c) \right| < \frac{\epsilon}{3}.$$By the mean value theorem applied to $f_n - f_m$ on $[c,x]$ (or $[x,c]$) with the above restriction on $x$ and $n,m \ge N$,$$\exists \xi \in (c,x): f_n^{\prime}(\xi) - f_m^{\prime}(\xi) = \frac{\left[f_n(x) - f_m(x) \right] - \left[f_n(c) - f_m(c) \right]}{x-c}$$Taking limit as $m \to \infty$, rearranging and substituting $n = N$, we get$$\left| \frac{f(x) - f(c)}{x-c} - \frac{f_N(x) - f_N(c)}{x-c} \right| < \frac{\epsilon}{3}$$by the second inequality above. Finally, we have$$\left| \frac{f(x) - f(c)}{x-c} - g(c) \right| =\left| \frac{f(x) - f(c)}{x-c} - \frac{f_N(x) - f_N(c)}{x-c} + \frac{f_N(x) - f_N(c)}{x-c} - {f_N}^{\prime}(c) + {f_N}^{\prime}(c) - g(c) \right| < \epsilon$$by the triangle inequality. Hence $f^{\prime}(c) = g(c)$ for any $c \in (a,b)$, i.e. $f^{\prime} = g$.

Question.At what point does this proof breaks if the convergence $f^{\prime} \to g$ is only pointwise? In such a case, $N_1$ and, consequently, $N$ and $\delta$ would depend on the point $c$ chosen, but I see no problem with that.

I know there are counter-examples to the similar claim with pointwise convergence, but it doesn't help. Also, I know this theorem can be proven alternatively via integrals, but I'd prefer to avoid using them.

I apologize if this post is too long. Any help will be appreciated.