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For $0\le t\le 1$, the triangle inequality givesTaking the supremum proves $\boxed{d(p_n,p)\le \rho (p_n,p)}$. Hence convergence for $\rho$ implies convergence for $d$.

The converse is false. For $n\ge 1$, take $q_n(t)=(1-t{)}^n/n$ and $p=0$. Its supremum norm is $d(q_n,0)=1/n\to 0$, whereas the binomial theorem givesThe alternating coefficients cancel on $[0,1]$; the polynomial coefficient sum norm measures their absolute values before any cancellation.

Extend the coefficient lists of each polynomial by zeros. Then $\rho (p,q)={\sum }_{k\ge 0}|a_k-b_k|$ is finite because only finitely many terms are nonzero. It is nonnegative and symmetric. Moreover, $\rho (p,q)=0$ holds exactly when every coefficient agrees, which by uniqueness of polynomial coefficients is equivalent to $p=q$.

If $r(t)=\sum c_k{t}^k$, the triangle inequality for the absolute value givesThus $\rho$ satisfies all the metric axioms. It is the metric induced by the polynomial coefficient sum norm.