= Solution
Every derivative of order $n+1$ of $\sin(2x)$ has absolute value at most $2^{n+1}$. The equally spaced nodes are symmetric about $1/2$. Pair each node $a\leq1/2$ with $1-a$; for $x\in[0,1]$,
$$
|(x-a)(x-(1-a))|=|(x-1/2)^2-(a-1/2)^2|\leq\frac14.
$$
If an unpaired middle node remains, its factor satisfies $|x-1/2|\leq1/2$. Consequently the complete nodal product is at most $2^{-(n+1)}$. The error formula actually gives the stronger estimate
$$
\|f-p_n\|_\infty\leq\frac1{(n+1)!}\leq\boxed{\frac{2^{\lfloor(n+1)/2\rfloor}}{(n+1)!}}.
$$
The first bound tends to zero, proving \b[<uniform convergence>] on the whole interval. There is no appeal to a general claim that equispaced interpolation always converges; the derivative and symmetric-product bounds are particular to this <analytic function> and these nodes.
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