[proofplan] We argue by contradiction: if no Lebesgue number exists, then for each $n \in \mathbb{N}$ the radius $1/n$ fails, providing a point $x_n \in X$ such that $B(x_n, 1/n)$ is not contained in any member of the cover. Compactness of $X$ yields a subsequence $x_{n_k} \to y$. Since $\mathcal{U}$ covers $X$, some $U_\alpha$ contains $y$ together with an open ball $B(y, r)$. For $k$ large, the triangle inequality forces $B(x_{n_k}, 1/n_k) \subseteq B(y, r) \subseteq U_\alpha$, contradicting the construction of the $x_n$. [/proofplan] [step:Assume no Lebesgue number exists and construct a sequence witnessing the failure] Suppose, for contradiction, that no $\delta > 0$ works. Then for every $\delta > 0$ there exists $x \in X$ such that $B(x, \delta) \not\subseteq U_\alpha$ for any $\alpha \in A$. Applying this with $\delta = 1/n$ for each $n \in \mathbb{N}$, we select a point \begin{align*} x_n \in X \quad \text{such that} \quad B(x_n, 1/n) \not\subseteq U_\alpha \text{ for any } \alpha \in A. \end{align*} This gives a sequence $(x_n)_{n \in \mathbb{N}}$ in $X$. The construction uses the axiom of dependent choice (or, if $A$ is indexed by a well-ordered set, ordinary choice), as is standard in metric-space arguments. [/step] [step:Extract a convergent subsequence using sequential compactness] Since $X$ is a compact metric space, it is sequentially compact by the [equivalence of compactness and sequential compactness in metric spaces](/theorems/???). Therefore the sequence $(x_n)$ has a convergent subsequence: there exist indices $n_1 < n_2 < \dots$ and a point $y \in X$ such that \begin{align*} x_{n_k} \to y \quad \text{in } (X, d) \text{ as } k \to \infty. \end{align*} Concretely, for every $\varepsilon > 0$ there exists $K \in \mathbb{N}$ with $d(x_{n_k}, y) < \varepsilon$ for all $k \geq K$. [guided] We have a sequence $(x_n)$ in $X$ and we want to extract a limit point. Compactness in a general topological space is defined via open covers, but metric spaces enjoy a stronger property: compactness is equivalent to sequential compactness, i.e., every sequence has a convergent subsequence. (This equivalence fails in general topological spaces — it relies on metric-space properties such as first countability.) Cite the [equivalence of compactness and sequential compactness in metric spaces](/theorems/???): the metric space $(X, d)$ is compact, so in particular every sequence in $X$ has a subsequence converging to a point of $X$. Apply this to $(x_n)$ to obtain the subsequence $x_{n_k} \to y$. Why do we pass to a subsequence rather than arguing about the original sequence? The original $(x_n)$ need not converge — it could oscillate around several limit points. Sequential compactness only guarantees that *some* subsequence converges, which is what we need for the triangle-inequality argument below. [/guided] [/step] [step:Locate $y$ in a member of the cover and produce an open ball around it] Since $\mathcal{U} = \{U_\alpha\}_{\alpha \in A}$ covers $X$, there exists some index $\alpha_0 \in A$ with $y \in U_{\alpha_0}$. Because $U_{\alpha_0}$ is open in $(X, d)$, by the definition of the metric topology there exists $r > 0$ such that \begin{align*} B(y, r) \subseteq U_{\alpha_0}. \end{align*} We fix this $r$ for the remainder of the argument. [/step] [step:Use the triangle inequality to force $B(x_{n_k}, 1/n_k)$ inside $B(y, r)$] Apply the convergence $x_{n_k} \to y$ with tolerance $r/2$: there exists $K_1 \in \mathbb{N}$ such that $d(x_{n_k}, y) < r/2$ for all $k \geq K_1$. Since $n_k \to \infty$ as $k \to \infty$, there exists $K_2 \in \mathbb{N}$ such that $1/n_k < r/2$ for all $k \geq K_2$. Fix any $k_0 \geq \max\{K_1, K_2\}$. We show \begin{align*} B(x_{n_{k_0}}, 1/n_{k_0}) \subseteq B(y, r). \end{align*} Let $z \in B(x_{n_{k_0}}, 1/n_{k_0})$, so $d(z, x_{n_{k_0}}) < 1/n_{k_0} < r/2$. By the triangle inequality, \begin{align*} d(z, y) \leq d(z, x_{n_{k_0}}) + d(x_{n_{k_0}}, y) < \frac{r}{2} + \frac{r}{2} = r, \end{align*} so $z \in B(y, r)$. Combining with $B(y, r) \subseteq U_{\alpha_0}$ from the previous step, \begin{align*} B(x_{n_{k_0}}, 1/n_{k_0}) \subseteq B(y, r) \subseteq U_{\alpha_0}. \end{align*} [guided] The strategy is to choose $k_0$ large enough that both the centre $x_{n_{k_0}}$ is close to $y$ (within $r/2$) *and* the radius $1/n_{k_0}$ is small (below $r/2$). The factor $1/2$ is chosen so that the two halves sum to exactly $r$ in the triangle inequality — a standard $\varepsilon/2$ trick. Why does this work? Any point $z$ in the smaller ball $B(x_{n_{k_0}}, 1/n_{k_0})$ is within $r/2$ of the centre $x_{n_{k_0}}$, which is within $r/2$ of $y$. The triangle inequality then puts $z$ within $r$ of $y$. Thus the smaller ball sits inside the ball $B(y, r)$, which in turn sits inside $U_{\alpha_0}$. Note that we need both conditions to hold simultaneously, which is why we take $k_0 \geq \max\{K_1, K_2\}$. Taking $k_0 \geq K_1$ alone would control the centre but not the radius; taking $k_0 \geq K_2$ alone would control the radius but not the centre. Compactness (via $K_1$) and the fact that $n_k \to \infty$ (via $K_2$) together furnish the common index. [/guided] [/step] [step:Derive the contradiction] By the construction in Step 1, the point $x_{n_{k_0}}$ was chosen precisely so that $B(x_{n_{k_0}}, 1/n_{k_0})$ is *not* contained in any member of $\mathcal{U}$. But we have just shown $B(x_{n_{k_0}}, 1/n_{k_0}) \subseteq U_{\alpha_0} \in \mathcal{U}$, a contradiction. This contradiction refutes the hypothesis that no Lebesgue number exists. Therefore there exists $\delta > 0$ such that for every $x \in X$, the ball $B(x, \delta)$ is contained in some $U_\alpha$. This completes the proof. [/step]