[proofplan] We first use compactness of $[0,1]$ to obtain a [uniform continuity](/page/Uniform%20Continuity) scale for $\gamma$. We then choose a uniform partition of $[0,1]$ whose mesh is smaller than that scale, and define $p$ by affine interpolation through the finitely many points $\gamma(t_i)$. On each subinterval, $p(t)$ is a convex combination of the two endpoint values of $\gamma$, so the triangle inequality bounds $|p(t)-\gamma(t)|$ by the corresponding endpoint errors. [/proofplan] [step:Choose a partition fine enough for uniform continuity] Since $\gamma:[0,1]\to\mathbb{R}^n$ is continuous and $[0,1]$ is compact, $\gamma$ is uniformly continuous. Hence there exists $\delta>0$ such that for all $s,t\in[0,1]$, \begin{align*} |s-t|<\delta \implies |\gamma(s)-\gamma(t)|<\varepsilon. \end{align*} Choose $m\in\mathbb{N}$ such that $1/m<\delta$. For each index $i\in\{0,\dots,m\}$, define \begin{align*} t_i:=\frac{i}{m}. \end{align*} Then \begin{align*} 0=t_0<t_1<\cdots<t_m=1, \end{align*} and each subinterval $[t_i,t_{i+1}]$, for $i\in\{0,\dots,m-1\}$, has length $1/m<\delta$. [guided] The only global input we need about the path is uniform continuity. Pointwise continuity would give a possibly different scale near each point, but a polygonal approximation is built from finitely many intervals, so we need one scale that works everywhere on $[0,1]$. Because $\gamma:[0,1]\to\mathbb{R}^n$ is continuous and the domain $[0,1]$ is compact, compactness upgrades continuity to uniform continuity. Thus, for the prescribed number $\varepsilon>0$, there exists a number $\delta>0$ such that whenever $s,t\in[0,1]$ satisfy $|s-t|<\delta$, the image points satisfy \begin{align*} |\gamma(s)-\gamma(t)|<\varepsilon. \end{align*} We now choose the partition so that any two points in the same subinterval are closer than $\delta$. Pick $m\in\mathbb{N}$ with $1/m<\delta$, and define the partition points by \begin{align*} t_i:=\frac{i}{m} \end{align*} for each $i\in\{0,\dots,m\}$. Then $0=t_0<t_1<\cdots<t_m=1$, and for every $i\in\{0,\dots,m-1\}$ the interval $[t_i,t_{i+1}]$ has length \begin{align*} t_{i+1}-t_i=\frac{1}{m}<\delta. \end{align*} Therefore, if $t\in[t_i,t_{i+1}]$, then both $|t-t_i|<\delta$ and $|t-t_{i+1}|<\delta$. [/guided] [/step] [step:Define the polygonal interpolant through the sampled points] Define $p:[0,1]\to\mathbb{R}^n$ as follows. For $t\in[t_i,t_{i+1}]$, where $i\in\{0,\dots,m-1\}$, let \begin{align*} \lambda_i(t):=\frac{t-t_i}{t_{i+1}-t_i}. \end{align*} Thus $\lambda_i(t)\in[0,1]$. Set \begin{align*} p(t):=(1-\lambda_i(t))\gamma(t_i)+\lambda_i(t)\gamma(t_{i+1}). \end{align*} This definition agrees at common endpoints because both adjacent formulas give $p(t_i)=\gamma(t_i)$. Hence $p$ is continuous on $[0,1]$ and affine on each subinterval $[t_i,t_{i+1}]$. Therefore $p$ is a polygonal path. [/step] [step:Estimate the interpolation error on each subinterval] Fix $t\in[0,1]$. Choose $i\in\{0,\dots,m-1\}$ such that $t\in[t_i,t_{i+1}]$, and write $\lambda:=\lambda_i(t)$. Since $0\leq\lambda\leq1$, the definition of $p$ gives \begin{align*} p(t)-\gamma(t)=(1-\lambda)(\gamma(t_i)-\gamma(t))+\lambda(\gamma(t_{i+1})-\gamma(t)). \end{align*} By the triangle inequality and homogeneity of the Euclidean norm, \begin{align*} |p(t)-\gamma(t)|\leq (1-\lambda)|\gamma(t_i)-\gamma(t)|+\lambda|\gamma(t_{i+1})-\gamma(t)|. \end{align*} Because $t,t_i\in[t_i,t_{i+1}]$ and $t,t_{i+1}\in[t_i,t_{i+1}]$, we have \begin{align*} |t-t_i|\leq t_{i+1}-t_i=\frac{1}{m}<\delta \end{align*} and \begin{align*} |t-t_{i+1}|\leq t_{i+1}-t_i=\frac{1}{m}<\delta. \end{align*} The choice of $\delta$ therefore implies \begin{align*} |\gamma(t_i)-\gamma(t)|<\varepsilon \end{align*} and \begin{align*} |\gamma(t_{i+1})-\gamma(t)|<\varepsilon. \end{align*} Substituting these two bounds into the previous inequality yields \begin{align*} |p(t)-\gamma(t)|<(1-\lambda)\varepsilon+\lambda\varepsilon=\varepsilon. \end{align*} Since this holds for every $t\in[0,1]$, it follows that \begin{align*} \sup_{t\in[0,1]} |p(t)-\gamma(t)|\leq\varepsilon. \end{align*} To obtain the strict supremum bound stated in the theorem, repeat the construction above with $\varepsilon/2$ in place of $\varepsilon$. Then the resulting polygonal path $p$ satisfies \begin{align*} |p(t)-\gamma(t)|<\frac{\varepsilon}{2} \end{align*} for every $t\in[0,1]$, and hence \begin{align*} \sup_{t\in[0,1]} |p(t)-\gamma(t)|\leq\frac{\varepsilon}{2}<\varepsilon. \end{align*} This is the desired polygonal approximation. [/step]