[proofplan] We argue by contradiction. If $f$ is continuous but not uniformly continuous on $[a,b]$, the negation of uniform continuity produces two sequences $(x_n)$ and $(y_n)$ with $|x_n - y_n| < 1/n$ yet $|f(x_n) - f(y_n)| \geq \varepsilon_0$. The Bolzano-Weierstrass theorem extracts a convergent subsequence of $(x_n)$, and the closeness condition forces $(y_n)$ to converge to the same limit. Continuity at this common limit then contradicts the separation of function values. [/proofplan] [step:Negate uniform continuity to obtain two sequences with separated function values] Suppose for contradiction that $f$ is not uniformly continuous on $[a,b]$. Then there exists $\varepsilon_0 > 0$ such that for every $\delta > 0$, there exist $x, y \in [a,b]$ with $|x - y| < \delta$ and $|f(x) - f(y)| \geq \varepsilon_0$. Applying this with $\delta = 1/n$ for each $n \in \mathbb{N}$ produces sequences $(x_n)$ and $(y_n)$ in $[a,b]$ satisfying \begin{align*} |x_n - y_n| < \frac{1}{n} \quad \text{and} \quad |f(x_n) - f(y_n)| \geq \varepsilon_0 \quad \text{for all } n \in \mathbb{N}. \end{align*} [/step] [step:Extract a convergent subsequence via Bolzano-Weierstrass and show both sequences share the same limit] Since $(x_n) \subset [a,b]$ and $[a,b]$ is closed and bounded, the Bolzano-Weierstrass theorem provides a subsequence $x_{n_k} \to c$ for some $c \in [a,b]$. The limit lies in $[a,b]$ because $[a,b]$ is closed. For the corresponding subsequence $(y_{n_k})$, the triangle inequality gives \begin{align*} |y_{n_k} - c| \leq |y_{n_k} - x_{n_k}| + |x_{n_k} - c| < \frac{1}{n_k} + |x_{n_k} - c|. \end{align*} Both terms on the right tend to zero as $k \to \infty$, so $y_{n_k} \to c$. [guided] We need to extract a convergent subsequence from $(x_n)$. The sequence lives in the closed bounded interval $[a,b]$, so Bolzano-Weierstrass applies directly: there exists a subsequence $(x_{n_k})$ converging to some $c \in \mathbb{R}$. Since $a \leq x_{n_k} \leq b$ for all $k$ and $[a,b]$ is closed, the limit $c$ belongs to $[a,b]$. The critical observation is that $(y_{n_k})$ must converge to the same limit $c$. We verify this using the triangle inequality: \begin{align*} |y_{n_k} - c| \leq |y_{n_k} - x_{n_k}| + |x_{n_k} - c| < \frac{1}{n_k} + |x_{n_k} - c|. \end{align*} The first term $1/n_k \to 0$ because $n_k \geq k \to \infty$. The second term $|x_{n_k} - c| \to 0$ by the definition of the subsequence. Therefore $y_{n_k} \to c$. The sequences are "squeezed together" by the condition $|x_n - y_n| < 1/n$, so any limit point of one is also a limit point of the other. [/guided] [/step] [step:Derive a contradiction from continuity at the common limit $c$] Since $f$ is continuous at $c \in [a,b]$, there exists $\delta_1 > 0$ such that $|t - c| < \delta_1$ implies $|f(t) - f(c)| < \varepsilon_0/2$. Choose $K$ large enough so that $|x_{n_K} - c| < \delta_1$ and $|y_{n_K} - c| < \delta_1$. The triangle inequality gives \begin{align*} |f(x_{n_K}) - f(y_{n_K})| \leq |f(x_{n_K}) - f(c)| + |f(c) - f(y_{n_K})| < \frac{\varepsilon_0}{2} + \frac{\varepsilon_0}{2} = \varepsilon_0. \end{align*} This contradicts $|f(x_{n_K}) - f(y_{n_K})| \geq \varepsilon_0$. The assumption that $f$ is not uniformly continuous is therefore false, and $f$ is uniformly continuous on $[a,b]$. [/step]