= Solution
Let $E=\mathbb R[x]_{\le d}$ with the <supremum norm>. By <equivalence of norms in finite dimensions>, there is $C_d$ such that $\sum_{j=0}^d|a_j|\le C_d\|p\|_\infty$ whenever $p=\sum_j a_jx^j$. Hence
$$
\|(B_n-I)p\|_\infty\le\sum_j|a_j|\|B_n(x^j)-x^j\|_\infty,
$$
so $\varepsilon_n:=\|B_n-I\|_{\rm op}\le C_d\max_{0\le j\le d}\|B_n(x^j)-x^j\|_\infty\to0$. This upgrades convergence on fixed polynomials to <operator norm> convergence on the fixed space.
For sufficiently large $n$, $\varepsilon_n<1$, and a <Neumann series> gives
$$
\|B_n^{-1}-I\|_{\rm op}\le\frac{\varepsilon_n}{1-\varepsilon_n}.
$$
Thus the <inverse Bernstein approximation on a fixed-degree polynomial space> satisfies $\|g_n-f\|_\infty\to0$. By the <extreme value theorem>, strict positivity on the compact interval gives $\mu=\min_{[0,1]}f>0$. Eventually $\|g_n-f\|_\infty<\mu/2$, and consequently
$$
\boxed{g_n(x)\ge\mu/2>0\quad\text{on }[0,1]\text{ for all sufficiently large }n.}
$$
Applying pointwise convergence to the varying polynomials $g_n$ without this finite-dimensional operator argument would not justify the conclusion.
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