[proofplan] We prove the result directly from continuity and compactness of the interval. Continuity of the path gives, around each point $a\in[0,1]$, a local interval on which the image under $\gamma$ has diameter less than $\varepsilon$ up to a triangle inequality. A [finite subcover](/page/Finite%20Subcover) of $[0,1]$ by these local intervals then lets us take the minimum of finitely many radii, producing one $\delta>0$ that works simultaneously for every pair $s,t\in[0,1]$. [/proofplan]

[step:Choose local continuity radii around every point of the interval]Let $\varepsilon>0$ be fixed. Since $\gamma:[0,1]\to X$ is a path, $\gamma$ is continuous at every $a\in[0,1]$. Therefore, for each $a\in[0,1]$, there exists a number $\rho_a>0$ such that for every $u\in[0,1]$, \begin{align*} |u-a|<\rho_a \implies d(\gamma(u),\gamma(a))<\frac{\varepsilon}{2}. \end{align*} For each $a\in[0,1]$, define the open interval $I_a\subset\mathbb{R}$ by \begin{align*} I_a:=\left(a-\frac{\rho_a}{2},a+\frac{\rho_a}{2}\right). \end{align*} Since $a\in I_a$ for every $a\in[0,1]$, the family $\{I_a:a\in[0,1]\}$ covers $[0,1]$.[/step]

[guided]Fix $\varepsilon>0$. The goal is to find one radius $\delta>0$ that works for all points of $[0,1]$. Continuity initially gives only pointwise radii, so we first record those local radii carefully. Because $\gamma:[0,1]\to X$ is a path, it is continuous at each point $a\in[0,1]$. Applying the $\varepsilon$-$\delta$ definition of continuity at $a$, with tolerance $\varepsilon/2$, gives a number $\rho_a>0$ such that every $u\in[0,1]$ satisfying $|u-a|<\rho_a$ also satisfies \begin{align*} d(\gamma(u),\gamma(a))<\frac{\varepsilon}{2}. \end{align*} We divide the radius by $2$ and define an open interval $I_a\subset\mathbb{R}$ by \begin{align*} I_a:=\left(a-\frac{\rho_a}{2},a+\frac{\rho_a}{2}\right). \end{align*} The factor $1/2$ leaves room for a second nearby point later: if $s$ lies within $\rho_a/2$ of $a$ and $t$ lies close enough to $s$, then $t$ will still lie within $\rho_a$ of $a$. Since $a\in I_a$ for every $a\in[0,1]$, the family $\{I_a:a\in[0,1]\}$ is an [open cover](/page/Open%20Cover) of $[0,1]$.[/guided]

[step:Pass from the open cover to finitely many local intervals] Using the compactness of the closed interval $[0,1]$, the open cover $\{I_a:a\in[0,1]\}$ has a finite subcover. Hence there exist an integer $m\in\mathbb{N}$ and points $a_1,\dots,a_m\in[0,1]$ such that \begin{align*} [0,1]\subset \bigcup_{i=1}^{m} I_{a_i}. \end{align*} Define \begin{align*} \delta:=\frac{1}{2}\min_{1\le i\le m}\rho_{a_i}. \end{align*} Since each $\rho_{a_i}>0$ and the minimum is taken over a finite nonempty set, $\delta>0$. [/step]

[step:Use the finite minimum to prove the uniform estimate] Let $s,t\in[0,1]$ satisfy $|s-t|<\delta$. Since the intervals $I_{a_1},\dots,I_{a_m}$ cover $[0,1]$, there exists an index $i\in\{1,\dots,m\}$ such that $s\in I_{a_i}$. By the definition of $I_{a_i}$, \begin{align*} |s-a_i|<\frac{\rho_{a_i}}{2}. \end{align*} Since $\delta\le \rho_{a_i}/2$, the triangle inequality in $\mathbb{R}$ gives \begin{align*} |t-a_i|\le |t-s|+|s-a_i|<\delta+\frac{\rho_{a_i}}{2}\le \rho_{a_i}. \end{align*} Therefore the defining property of $\rho_{a_i}$ gives \begin{align*} d(\gamma(s),\gamma(a_i))<\frac{\varepsilon}{2} \end{align*} and \begin{align*} d(\gamma(t),\gamma(a_i))<\frac{\varepsilon}{2}. \end{align*} Applying the triangle inequality in the [metric space](/page/Metric%20Space) $(X,d)$, \begin{align*} d(\gamma(s),\gamma(t))\le d(\gamma(s),\gamma(a_i))+d(\gamma(a_i),\gamma(t))<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \end{align*} Thus for this $\delta>0$, every $s,t\in[0,1]$ with $|s-t|<\delta$ satisfy $d(\gamma(s),\gamma(t))<\varepsilon$. Since $\varepsilon>0$ was arbitrary, $\gamma$ is uniformly continuous on $[0,1]$. [/step]