Take an infinity norm defined as $\|f\|_\infty=\max_{x\in[a,b]}|f(x)|$. This is basically an $L_\infty$ norm. Also let $q_n$ be a polynomial of degree $n$ and $c_i,i=0,\ldots,n$ be its $n+1$ coefficients.
Also define a function $E(c_0,c_1,\ldots,c_n)=\|f-q_n\|_\infty$. If $S=\{\textbf{c}=(c_0,c_1,\ldots,c_n)\in\mathbb{R^{n+1}}:E(\textbf{c})\le\|f\|_\infty+1\}$. How to show that this set $S$ is compact?
My idea is to show that this set is closed and bounded since that will imply compactness.
Just to clarify, letting $\phi_c(x) = \sum_k c_k x^k$, you are asking if $S=\{ c | \|f-\phi_c\|_\infty \le 1+ \|f\|_\infty\}$ is compact.
Since $S$ is finite dimensional, we just need to show that $S$ is closed and bounded.
Note that $\|c\|_* = \|\phi_c\|_\infty$ is a norm on $\mathbb{R}^{n+1}$, and since the space is finite dimensional, all norms are equivalent, in particular, $\|\cdot\|_*$ is continuous.
If $c \in S$ then $\|c\|_* = \|\phi_c\|_\infty \le 1+2 \|f\|_\infty$. Hence $S$ is bounded.
Since $c \mapsto \|f-\phi_c\|_\infty$ is continuous, $S$ is closed.