My proof: Let $\varepsilon > 0$ be arbitrary. Since $f$ is continuous on the compact set $[a, b]$, $f$ is uniformly continuous on $[a, b]$. Thus there exists $\delta > 0$ such that for each $x, y \in [a, b]$, $|x - y| < \delta$ implies $|f(x) - f(y)| < \frac{\varepsilon}{2}$. Now partition $[a, b]$ into $k = \text{ceil}\left(\frac{b - a}{\frac{\delta}{2}}\right)$ intervals $[t_1 = a, t_2), [t_2, t_3), \dots, [t_{k - 1}, t_k), [t_k, t_{k + 1} = b]$ with $t_{j + 1} - t_j \leq \frac{\delta}{2}$ for each $j \in \{1, 2, \dots, k\}$. For each $j \in \{1, 2, \dots, k\}$, let $N_j$ be such that for each $n \geq N_j$, $|f_n(t_j) - f(t_j)| < \frac{\varepsilon}{2}$. Let $N = \max(N_1, N_2, \dots, N_k)$. Now to show uniform convergence, let $n \geq N, x \in [a, b]$ be arbitrary. Let $t_i$ be the left endpoint of the interval containing $x$. Since $f_n$ is increasing and $t_i \leq x \leq t_{i + 1}$, we have $f_n(t_i) \leq f_n(x) \leq f_n(t_{i + 1})$. Thus we have $f_n(x) - f(x) \geq f_n(t_i) - f(x) = f_n(t_i) - f(t_i) + f(t_i) - f(x) > -\frac{\varepsilon}{2} + -\frac{\varepsilon}{2} = -\varepsilon$ and $f_n(x) - f(x) \leq f_n(t_{i + 1}) - f(x) = f_n(t_{i + 1}) - f(t_{i + 1}) + f(t_{i + 1}) - f(x) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$. Thus $|f_n(x) - f(x)| < \varepsilon$. Thus $(f_n) \to f$ uniformly.
Nice proof. It looks fine to me and I didn't find any mistakes. (I would leave this as a comment instead but I do not have enough reputation points)