Let $(X,\|.\|)$ be the normed vector space of polynomials with $\|p_0+p_1t+\dots+p_nt^n\|=\max\{|p_0|,\dots,|p_n|\}$ with normal addition and multiplication by a scalar.
I have proved that $U_n=\{P(t) :\deg P>n\}$ is dense in $(X,\|.\|)=$polynomials over $\Bbb R$ with the $\max|p_i|$ norm:
for any $P\in X$ and $\delta>0$ if $\deg P>n$ then $P\in U_n\cap B(P,\delta)\ne\emptyset$
If $\deg P\le n$ then $Q:=P+{\delta\over2}t^{n+1}$ is in the ball $B(P,\delta)$ and in $U_n$ simultaneously so $U_n$ is dense since $P$ and $\delta$ were arbitrary
But for proving that $U_n$ is open for any $n\in\Bbb N$ I'm not sure if this is ok:
Let $P\in U_n$ and $d:=\deg P$. Let $r:=\min\{|p_i|:i\in\{n,\dots,d\}, p_i\ne 0\}$
If $Q\in B(P,r)$ then if $\deg Q\le n$, $\|P-Q\|\ge\|p_{n+1}t^{n+1}+p_{n+2}t^{n+2}+\dots+p_dt^d\|\ge r$
Does that make sense?
I don't know where does the inequality $\lVert P-Q\rVert\geqslant\lVert p_nt^n+p_{n+1}t^{n+1}+\dots+p_dt^d\rVert\geqslant r$ came from (or what it means).
Note that, by the definition $\lVert\cdot\rVert$, if $P(x)=p_Nx^N+p_{N-1}x^{N-1}+\cdots+p_0\in U_n$, and $Q(x)\in B\bigl(P(x),\lvert p_N\rvert\bigr)$, then $Q(x)\in U_n$. Therefore, $U_n$ is an open set.