= Solution
Here
$$
d_1(f,g)=\int_0^1|f(x)-g(x)|\,dx,
\qquad
d_\infty(f,g)=\max_{x\in[0,1]}|f(x)-g(x)|.
$$
The <L1 norm> and <uniform norm> satisfy
$$
d_1(f,g)\le d_\infty(f,g),
$$
so the identity from the $d_\infty$ metric to the $d_1$ metric is one-Lipschitz and therefore continuous. The two metrics do not induce the same topology on all of $C[0,1]$. For
$$
f_n(x)=\max(1-nx,0),
$$
one has $d_1(f_n,0)=1/(2n)\to0$ but $d_\infty(f_n,0)=1$. Thus $d_1$ convergence need not imply uniform convergence.
A map $F:(X,d_X)\to(Y,d_Y)$ is <Lipschitz continuous> if there is $L<\infty$ such that
$$
d_Y(F(x),F(x'))\le Ld_X(x,x')
$$
for all $x,x'\in X$. Evaluation at $a\in[0,1]$ is one-Lipschitz because
$$
|f(a)-g(a)|\le d_\infty(f,g).
$$
Choose distinct $a_0,\ldots,a_n\in[0,1]$ and define
$$
T:\mathcal P_n\longrightarrow\mathbb R^{n+1},
\qquad
T(p)=(p(a_0),\ldots,p(a_n)).
$$
The <Vandermonde determinant> is nonzero, so $T$ is a bijection. It is one-Lipschitz for the two uniform metrics. If $\ell_i$ are the associated <Lagrange interpolation polynomial>[Lagrange cardinal polynomials], then
$$
p(x)=\sum_{i=0}^np(a_i)\ell_i(x),
$$
and hence
$$
\|p-q\|_\infty
\le\left(\sum_{i=0}^n\|\ell_i\|_\infty\right)
\|T(p)-T(q)\|_\infty.
$$
Thus $T^{-1}$ is also Lipschitz.
Let $\widehat{\mathcal P}_n$ be the polynomials whose values lie in $[-1,1]$. Its image under $T$ lies in the bounded cube $[-1,1]^{n+1}$. It is closed: if $T(p_j)\to v$, then the Lipschitz inverse gives uniform convergence $p_j\to p=T^{-1}(v)$, and passing to the limit pointwise preserves $|p(x)|\le1$. By the <Heine-Borel theorem>, $T(\widehat{\mathcal P}_n)$ is compact, and the continuous inverse carries compactness back to $\widehat{\mathcal P}_n$. Therefore
$$
\boxed{(\widehat{\mathcal P}_n,d_\infty)\text{ is compact}}.
$$
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