[proofplan] We prove continuity at an arbitrary point $a \in X$. Given an error tolerance $\varepsilon>0$, [uniform continuity](/page/Uniform%20Continuity) provides a single distance tolerance $\delta>0$ that works for every pair of points in $X$. Applying that global condition to the particular pair $(a,z)$ gives exactly the $\varepsilon$-$\eta$ condition for continuity at $a$. Since $a$ was arbitrary, $f$ is continuous on all of $X$. [/proofplan] [step:Fix an arbitrary point and an error tolerance] Let $a \in X$ be arbitrary, and let $\varepsilon \in \mathbb{R}$ satisfy $\varepsilon>0$. To prove that $f$ is continuous at $a$, it is enough to find a real number $\eta>0$ such that for every $z \in X$, the implication $d_X(a,z)<\eta \implies d_Y(f(a),f(z))<\varepsilon$ holds. [/step] [step:Choose the uniform continuity tolerance for the same error] Since $f:X \to Y$ is uniformly continuous, applying the definition of uniform continuity to this $\varepsilon>0$ gives a real number $\delta>0$ such that for all $u,v \in X$, $d_X(u,v)<\delta \implies d_Y(f(u),f(v))<\varepsilon$. [guided] The local definition of continuity at $a$ asks for a distance tolerance that controls $d_Y(f(a),f(z))$ whenever $z$ is close to $a$. Uniform continuity gives something stronger: for the same fixed $\varepsilon>0$, it provides one real number $\delta>0$ that works simultaneously for every ordered pair of points in $X$. Precisely, because $f:X \to Y$ is uniformly continuous and $\varepsilon>0$, there exists $\delta>0$ such that for every $u \in X$ and every $v \in X$, $d_X(u,v)<\delta \implies d_Y(f(u),f(v))<\varepsilon$. This statement is stronger than what continuity at $a$ requires because the points $u$ and $v$ are not tied to any particular base point. We will use it only in the special case $u=a$ and $v=z$. [/guided] [/step] [step:Specialize the global implication to pairs based at the fixed point] Set $\eta := \delta$. Then $\eta>0$. Let $z \in X$ satisfy $d_X(a,z)<\eta$. Since $\eta=\delta$, we have $d_X(a,z)<\delta$. Applying the uniform continuity implication with $u=a$ and $v=z$ gives $d_Y(f(a),f(z))<\varepsilon$. Thus $f$ is continuous at $a$. [/step] [step:Conclude continuity on the whole domain] The point $a \in X$ was arbitrary. Therefore $f$ is continuous at every point of $X$, and hence $f$ is continuous on $X$. [/step]