[proofplan] We choose a minimizing sequence in $V$ whose distances to $f$ converge to the infimum $E_V(f)_X$. The triangle inequality shows that this sequence is bounded in the finite-dimensional normed space $V$. By compactness of closed bounded sets in finite-dimensional normed spaces, a subsequence converges in $V$ to some $v^* \in V$. Continuity of the norm then passes the limiting distance through the subsequence and proves that $v^*$ attains the infimum. [/proofplan] [step:Choose a minimizing sequence for the best approximation error] Define the real number \begin{align*} e := E_V(f)_X = \inf_{v \in V} \|f - v\|_X. \end{align*} Since $V$ is a linear subspace, $0 \in V$, so $0 \le e \le \|f\|_X < \infty$. For each $k \in \mathbb{N}$, the definition of infimum gives an element $v_k \in V$ such that \begin{align*} \|f - v_k\|_X \le e + \frac{1}{k}. \end{align*} Thus $(v_k)_{k \in \mathbb{N}}$ is a sequence in $V$ satisfying \begin{align*} \lim_{k \to \infty} \|f - v_k\|_X = e. \end{align*} Indeed, the lower bound $e \le \|f - v_k\|_X$ holds for every $k \in \mathbb{N}$ by definition of $e$, while the upper bound above tends to $e$. [/step] [step:Bound the minimizing sequence inside a closed ball of $V$] For every $k \in \mathbb{N}$, the triangle inequality in $X$ gives \begin{align*} \|v_k\|_V = \|v_k\|_X \le \|v_k - f\|_X + \|f\|_X. \end{align*} Using the minimizing-sequence estimate, \begin{align*} \|v_k\|_V \le e + \frac{1}{k} + \|f\|_X \le e + 1 + \|f\|_X. \end{align*} Define \begin{align*} R := e + 1 + \|f\|_X. \end{align*} Then $R > 0$ and $v_k$ belongs to the closed ball \begin{align*} K := \{v \in V : \|v\|_V \le R\} \end{align*} for every $k \in \mathbb{N}$. [/step] [step:Use finite-dimensional compactness to extract a convergent subsequence] The set $K$ is closed and bounded in the finite-dimensional normed space $(V,\|\cdot\|_V)$. By the finite-dimensional Heine-Borel [compactness theorem](/theorems/2748) (citing a result not yet in the wiki: closed bounded subsets of finite-dimensional normed spaces are compact), $K$ is compact. Therefore the sequence $(v_k)_{k \in \mathbb{N}}$ has a subsequence $(v_{k_j})_{j \in \mathbb{N}}$ and an element $v^* \in K$ such that \begin{align*} \lim_{j \to \infty} \|v_{k_j} - v^*\|_V = 0. \end{align*} Since $K \subset V$, we have $v^* \in V$. [guided] The purpose of the preceding boundedness estimate is to place the whole minimizing sequence inside one compact set. We defined \begin{align*} K := \{v \in V : \|v\|_V \le R\}, \end{align*} where $R = e + 1 + \|f\|_X$, and proved that $v_k \in K$ for every $k \in \mathbb{N}$. Now we use the finite-dimensional hypothesis. The finite-dimensional Heine-Borel compactness theorem states that every closed and bounded subset of a finite-dimensional normed space is compact. This is the only point where finite-dimensionality is essential. The set $K$ is bounded by construction, because every $v \in K$ satisfies $\|v\|_V \le R$. It is closed in $V$ because it is the inverse image of the closed interval $[0,R] \subset \mathbb{R}$ under the norm map $v \mapsto \|v\|_V$, and the norm map is continuous. Hence $K$ satisfies the hypotheses of the compactness theorem. Therefore $K$ is compact. Since $(v_k)_{k \in \mathbb{N}}$ is a sequence in the compact [metric space](/page/Metric%20Space) $K$, it has a convergent subsequence. Thus there exist a strictly increasing sequence of indices $(k_j)_{j \in \mathbb{N}}$ in $\mathbb{N}$ and an element $v^* \in K$ such that \begin{align*} \lim_{j \to \infty} \|v_{k_j} - v^*\|_V = 0. \end{align*} Because $K$ is a subset of $V$, the [limit point](/page/Limit%20Point) $v^*$ belongs to $V$. [/guided] [/step] [step:Pass the distance to the limit and identify the minimizer] Since the norm on $V$ is inherited from $X$, convergence in $\|\cdot\|_V$ means \begin{align*} \lim_{j \to \infty} \|v_{k_j} - v^*\|_X = 0. \end{align*} For every $j \in \mathbb{N}$, the [reverse triangle inequality](/theorems/2300) in $X$ gives \begin{align*} \left| \|f - v_{k_j}\|_X - \|f - v^*\|_X \right| \le \|v_{k_j} - v^*\|_X. \end{align*} Taking $j \to \infty$ yields \begin{align*} \lim_{j \to \infty} \|f - v_{k_j}\|_X = \|f - v^*\|_X. \end{align*} But $(v_{k_j})_{j \in \mathbb{N}}$ is a subsequence of the minimizing sequence, so \begin{align*} \lim_{j \to \infty} \|f - v_{k_j}\|_X = e. \end{align*} Therefore \begin{align*} \|f - v^*\|_X = e = E_V(f)_X. \end{align*} Since $v^* \in V$, this proves that $v^*$ is a best approximation to $f$ from $V$. [/step]