= Solution
Interpret convergence in $C[0,1]$ as <uniform convergence>, and work first with real <continuous functions>. Set $e_{n,j}=U_n(p_j)-p_j$, for $j=0,1,2$, and put $M=\|f\|_\infty$. The given barriers have one $\gamma\ge0$ independent of both $t$ and $n$. A <positive linear operator on continuous functions> preserves non-strict inequalities, so application of $U_n$ and evaluation at $t$ give
$$
\left|U_nf(t)-f(t)U_n1(t)\right|\le\varepsilon U_n1(t)+\gamma U_n((\cdot-t)^2)(t).
$$
Strict pointwise inequalities need not remain strict under a positive operator; non-strict ones suffice. Linearity and the cancellation of the exact moments yield
$$
U_n((\cdot-t)^2)(t)=e_{n,2}(t)-2t e_{n,1}(t)+t^2e_{n,0}(t)\ge0.
$$
Subtracting $f(t)$ rather than $f(t)U_n1(t)$ therefore gives the <quadratic barrier estimate for positive approximation operators>:
$$
\|U_nf-f\|_\infty\le\varepsilon\bigl(1+\|e_{n,0}\|_\infty\bigr)+M\|e_{n,0}\|_\infty+\gamma\bigl(\|e_{n,2}\|_\infty+2\|e_{n,1}\|_\infty+\|e_{n,0}\|_\infty\bigr).
$$
For fixed $\varepsilon$, all three moment errors tend uniformly to zero. Thus $\limsup_n\|U_nf-f\|_\infty\le\varepsilon$. Since $\varepsilon$ was arbitrary,
$$
\boxed{\|U_nf-f\|_\infty\longrightarrow0.}
$$
This completes the <Korovkin theorem> proof. Complex-valued functions follow by applying the real result to real and imaginary parts. <Uniform convergence> of the three test functions is essential to this uniform conclusion; purely pointwise test convergence should not be silently substituted.
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