The idea of the proof is to approximate $\frac{1}{x}$ in some way by polynomials $\pi_k$ so that $x \mapsto x \pi_k(x)$ is approximately constant, and the integral behaves like $\int \frac{1}{x} dx$.
Define the continuous function $f_\delta(x) = \frac{1}{\delta}1_{[0,\delta ]}(x) + \frac{1}{x} 1_{(\delta,\infty)}(x)$, with $0<\delta < 1$, and choose $N=1$. Let $\pi_k$ be a sequence of polynomials that converge uniformly to $f_\delta$ on $[0,1]$; define the polynomial $p_k$ by $p_k(x) = x \pi_k(x)$, and note that $|p_k(x) - x f_\delta(x)| \leq x \|f_\delta - \pi_k \|_\infty$, for $x \in [0,1]$.
Suppose $\|f_\delta - \pi_k \|_\infty < 1$, then $\frac{\sin p_k(x)}{x} = \frac{\sin x f_\delta(x)}{x} + \frac{\sin x \pi_k(x)-\sin x f_\delta(x)}{x} \geq \frac{\sin x f_\delta(x)}{x} - 1$, using the fact that $\sin$ is Lipschitz with rank 1. Consequently, $\int_0^1 \frac{\sin p_k(x)}{x} dx \geq \int_0^1 \frac{\sin x f_\delta(x)}{x} dx -1$.
Then we have $\int_0^1 \frac{\sin x f_\delta(x)}{x} dx = \int_0^\delta \frac{\sin \frac{x}{\delta} }{x} dx + \int_\delta^1 \frac{\sin 1}{x} dx \geq \frac{1}{2} +\sin 1 ( 0 - \ln \delta ) \geq \sin 1 \ln \frac{1}{\delta}$, where we use the fact that $\sin x \geq \frac{x}{2}$ for $x \in [0,1]$.
Hence, for any $\delta>0$, there exists a $k$ such that $\int_0^1 \frac{\sin p_k(x)}{x} dx \geq \sin 1 \ln \frac{1}{\delta} -1$, and since the right hand side is unbounded, this shows that no $C$ exists satisfying the above inequality.