Proving uniform convergence of a Fourier series

$\def\R{\mathbb{R}}\def\N{\mathbb{N}}\def\paren#1{\left(#1\right)}$The counter-example provided by @Conrad is enough to disprove the proposition. A stronger proposition, however, can be proved:

If $f: [-π, π] → \R$ is strictly monotonic and has only a finite number of points of discontinuity, and $g$ is a $2π$-periodic function with $f = g$ on $(-π, π)$, then the Fouries series of $g$ does not converge uniformly on $(0, π)$.

Proof: Denote the partial sums of the Fourier series by $S_N(x) = a_0 + \sum\limits_{n = 1}^N (a_n \cos nx + b_n \sin nx)$ for $N \in \N$. Note that the Dirichlet condition implies that$$ \lim_{N → ∞} S_N(x) = \frac{1}{2} (g(x + 0) + g(x - 0)).\quad \forall x \in \R $$ Suppose $\{S_N\}$ converges uniformly on $(0, π)$, then\begin{gather*} \frac{1}{2} (g(π + 0) + g(π - 0)) = \lim_{N → ∞} S_N(π) = \lim_{N → ∞} \lim_{x → π-} S_N(x)\\ \stackrel{(1)}{=} \lim_{x → π-} \lim_{N → ∞} S_N(x) = \lim_{x → π-} \frac{1}{2} (g(x + 0) + g(x - 0)) \stackrel{(2)}{=} g(π - 0), \end{gather*} where (1) uses uniform convergence and (2) uses monotonicity of $g$ on $(-π, π)$. But$$ g(π - 0) = f(π - 0) \geqslant f\paren{ \frac{π}{3} } > f\paren{ -\frac{π}{3} } \geqslant f(-π + 0) = g(-π + 0) = g(π + 0),$$ a contradiction.