[proofplan]
Since $K$ is compact, every [continuous function](/page/Continuous%20Function) on $K$ has finite uniform norm. We restrict the minimisation problem to the set of approximants whose error is no worse than the zero approximant. This set is closed and bounded in the finite-dimensional normed space $V$, hence compact. The error functional $v \mapsto \|f-v\|_\infty$ is continuous on $V$, so it attains a minimum on this compact set, and the restriction does not change the global infimum over $V$.
[/proofplan]
[step:Reduce the minimisation to approximants no worse than the zero function]
Fix $f \in C(K)$. Since $K$ is compact and $f: K \to \mathbb{R}$ is continuous, $f$ is bounded on $K$, so $\|f\|_\infty < \infty$. Since $V$ is a linear subspace of $C(K)$, the zero function $0_K: K \to \mathbb{R}$ belongs to $V$. Define the real number
\begin{align*}
R := \|f - 0_K\|_\infty = \|f\|_\infty.
\end{align*}
Define the admissible comparison set $A \subset V$ by
\begin{align*}
A := \{v \in V : \|f-v\|_\infty \le R\}.
\end{align*}
The set $A$ is nonempty because $0_K \in A$.
If $v \in V \setminus A$, then $\|f-v\|_\infty > R = \|f-0_K\|_\infty$, so such a $v$ cannot improve on the zero approximant. Therefore
\begin{align*}
\inf_{v \in V} \|f-v\|_\infty = \inf_{v \in A} \|f-v\|_\infty.
\end{align*}
[guided]
The minimisation is initially over all of $V$, which need not be compact. To force compactness, we discard every approximant whose error is already worse than the error of the zero function.
The compactness of $K$ is used here to ensure that the uniform norm of $f$ is finite: since $f: K \to \mathbb{R}$ is continuous on a compact [metric space](/page/Metric%20Space), $f$ is bounded, and therefore $\|f\|_\infty < \infty$. Because $V$ is a linear subspace, it contains the zero function $0_K: K \to \mathbb{R}$. We use it as a fixed comparison element and define
\begin{align*}
R := \|f - 0_K\|_\infty = \|f\|_\infty.
\end{align*}
Now define
\begin{align*}
A := \{v \in V : \|f-v\|_\infty \le R\}.
\end{align*}
This set is nonempty, since $0_K \in V$ and $\|f-0_K\|_\infty = R$.
Why is it legitimate to minimise only over $A$? If $v \in V \setminus A$, then by definition of $A$ we have
\begin{align*}
\|f-v\|_\infty > R = \|f-0_K\|_\infty.
\end{align*}
Thus $v$ is strictly worse than the already available approximant $0_K$. Such an element cannot be a best approximant and cannot lower the infimum. Hence the infimum over all of $V$ equals the infimum over $A$:
\begin{align*}
\inf_{v \in V} \|f-v\|_\infty = \inf_{v \in A} \|f-v\|_\infty.
\end{align*}
[/guided]
[/step]
[step:Show that the restricted admissible set is compact]
Define the error functional $\Phi: V \to \mathbb{R}$ by
\begin{align*}
\Phi(v) := \|f-v\|_\infty.
\end{align*}
The space $V$ is a finite-dimensional normed space when equipped with the inherited uniform norm from $C(K)$. The map $\Phi$ is continuous on this normed space because, for all $v,w \in V$, the [reverse triangle inequality](/theorems/2300) gives
\begin{align*}
|\Phi(v)-\Phi(w)| \le \|(f-v)-(f-w)\|_\infty = \|v-w\|_\infty.
\end{align*}
Therefore $A = \Phi^{-1}((-\infty,R])$ is closed in $V$.
The set $A$ is bounded in $V$. Indeed, if $v \in A$, then the triangle inequality gives
\begin{align*}
\|v\|_\infty \le \|v-f\|_\infty + \|f\|_\infty \le R + R = 2R.
\end{align*}
Thus $A$ is closed and bounded as a subset of the finite-dimensional normed space $(V,\|\cdot\|_\infty)$. By the [Heine-Borel theorem](/theorems/309) for finite-dimensional normed spaces, a closed and bounded subset of a finite-dimensional normed space is compact, so $A$ is compact.
[/step]
[step:Minimize the error functional on the compact admissible set]
Since $A$ is compact and the restriction $\Phi|_A: A \to \mathbb{R}$ is continuous, the [Extreme Value Theorem for Compact Spaces](/theorems/304) implies that $\Phi|_A$ attains its minimum on $A$. Hence there exists $v_* \in A$ such that
\begin{align*}
\Phi(v_*) = \min_{v \in A} \Phi(v).
\end{align*}
Equivalently,
\begin{align*}
\|f-v_*\|_\infty = \min_{v \in A} \|f-v\|_\infty.
\end{align*}
Using the reduction from the first step, this minimum is the global infimum over $V$:
\begin{align*}
\|f-v_*\|_\infty = \inf_{v \in V} \|f-v\|_\infty.
\end{align*}
Therefore $v_*$ is a best approximation to $f$ from $V$.
[guided]
At this point the problem has become a standard compactness argument. The previous step proved that $A$ is compact, and the same Lipschitz estimate used there proves that $\Phi|_A: A \to \mathbb{R}$ is continuous. Therefore the hypotheses of the [Extreme Value Theorem for Compact Spaces](/theorems/304) are satisfied: the domain $A$ is compact, the codomain is $\mathbb{R}$, and the function $\Phi|_A$ is continuous.
The theorem gives an element $v_* \in A$ at which the minimum is attained:
\begin{align*}
\Phi(v_*) = \min_{v \in A} \Phi(v).
\end{align*}
By the definition of $\Phi$, this is the same as
\begin{align*}
\|f-v_*\|_\infty = \min_{v \in A} \|f-v\|_\infty.
\end{align*}
Finally, the first step proved that restricting from all of $V$ to $A$ does not change the infimum. Hence
\begin{align*}
\|f-v_*\|_\infty = \inf_{v \in V} \|f-v\|_\infty.
\end{align*}
Since $v_* \in A \subset V$, this says exactly that $v_*$ is a best approximation to $f$ from $V$.
[/guided]
[/step]
