Show there is an $n$-th degree polynomial $p(x)$ such that $||f(x)-p(x)||_\infty \leq ||f(x)-q(x)||_\infty$. Show that for each $f \in C[0,1]$ there is an $n$-th degree polynomial $p(x)$ on $[0,1]$ such that $||f(x)-p(x)||_\infty \leq ||f(x)-q(x)||_\infty$ for any other $n$-th degree polynomial $q(x)$.
This looks similar to the following

If $A \subseteq (X,||\cdot||)$ is compact and non-empty then for each $x \in X$ there is some $y_0 \in A$ such that
  $$||x-y_0||=\inf\{||x-y||: y \in A\}$$

However, the set of $n$-th degree polynomial is finite-dimensional (hence closed), can we prove the it is compact?
 A: 
Theorem (Distance between compact sets): Let $(X,d)$ be some metric space, $A,B\subset X$ be nonempty and compact.
  Then there are $x\in A$ and $y\in B$ with
  $$d(x,y)=d(A,B).$$

Here, the distance between two sets $P,Q\subset X$ is defined as
$$d(P,Q) = \inf_{x\in P, y\in Q}d(x,y).$$
Proof: Let $(x_n,y_n)_{n\in\Bbb N}\subset A\times B$ be a minimizing sequence, say
$$d(A,B) \leqslant d(x_n,y_n) < d(A,B)+\frac1n.$$
Because $A$ is compact, there is a subsequence $(x_{n_k})_{k\in\mathbb N}$ that is convergent, say to $x\in A$.
Because $B$ is compact, there is a subsequence $(y_{n_{k_j}})_{j\in\mathbb N}$ that is convergent, say to $y\in B$.
Then, as $j\to\infty$ we see from
$$d(A,B) \leqslant d\left(x_{n_{k_j}},y_{n_{k_j}}\right) < d(A,B)+\frac1{{n_{k_j}}}$$
that $d(x,y) = d(A,B)$. $\square$

We will use the theorem with $A = \{f\}\subset C[0,1]$.
For $B$, let $P_n\subset C[0,1]$ be the subset of $n$-degree polynomials.
Pick any $q\in P_n$ and consider $B = \overline{\mathcal B(f,\lVert q\rVert)}\cap P_n$.
It suffices to check that $B$ is compact.
To that end, notice that $P_n$ is a closed, finite-dimensional subspace of $C[0,1]$, so that $B$ is the intersection of two closed sets and hence closed.
Moreover, $B$ is by construction bounded, so it is a closed and bounded subset of a finite-dimensional normed space, and hence compact.
The theorem will produce some $p$ that minimizes distance to $f$ over $B$.
Can you see why $p$ minimizes distance over all of $P_n$? (Use the triangle inequality!)
