Direct Proof

Let ${\displaystyle {\mathcal {U}}}$ be an open cover of ${\displaystyle X}$. Since ${\displaystyle X}$ is compact we can extract a finite subcover ${\displaystyle \{A_{1},\dots ,A_{n}\}\subseteq {\mathcal {U}}}$. If any one of the ${\displaystyle A_{i}}$'s equals ${\displaystyle X}$ then any ${\displaystyle \delta >0}$ will serve as a Lebesgue number. Otherwise for each ${\displaystyle i\in \{1,\dots ,n\}}$, let ${\displaystyle C_{i}:=X\smallsetminus A_{i}}$, note that ${\displaystyle C_{i}}$ is not empty, and define a function ${\displaystyle f:X\rightarrow \mathbb {R} }$ by

${\displaystyle f(x):={\frac {1}{n}}\sum _{i=1}^{n}d(x,C_{i}).}$

Since ${\displaystyle f}$ is continuous on a compact set, it attains a minimum ${\displaystyle \delta }$. The key observation is that, since every ${\displaystyle x}$ is contained in some ${\displaystyle A_{i}}$, the extreme value theorem shows ${\displaystyle \delta >0}$. Now we can verify that this ${\displaystyle \delta }$ is the desired Lebesgue number. If ${\displaystyle Y}$ is a subset of ${\displaystyle X}$ of diameter less than ${\displaystyle \delta }$, then there exists ${\displaystyle x_{0}\in X}$ such that ${\displaystyle Y\subseteq B_{\delta }(x_{0})}$, where ${\displaystyle B_{\delta }(x_{0})}$ denotes the ball of radius ${\displaystyle \delta }$ centered at ${\displaystyle x_{0}}$ (namely, one can choose ${\displaystyle x_{0}}$ as any point in ${\displaystyle Y}$). Since ${\displaystyle f(x_{0})\geq \delta }$ there must exist at least one ${\displaystyle i}$ such that ${\displaystyle d(x_{0},C_{i})\geq \delta }$. But this means that ${\displaystyle B_{\delta }(x_{0})\subseteq A_{i}}$ and so, in particular, ${\displaystyle Y\subseteq A_{i}}$.

Proof by Contradiction

Assume ${\displaystyle X}$ is sequentially compact, ${\displaystyle {\mathcal {A}}=\{U_{\alpha }|\alpha \in J\}}$ is an open covering of ${\displaystyle X}$ and the Lebesgue number ${\displaystyle \delta }$ does not exist. So, ${\displaystyle \forall \delta >0}$, ${\displaystyle \exists A\subset X}$ with ${\displaystyle diam(A)<\delta }$ such that ${\displaystyle \neg \exists \beta \in J}$ where ${\displaystyle A\subset U_{\beta }}$.

This allows us to make the following construction:

${\displaystyle \delta _{1}=1}$, ${\displaystyle \exists A_{1}\subset X}$ where ${\displaystyle (diam(A_{1})<\delta _{1})}$ and ${\displaystyle \neg \exists \beta (A_{1}\subset U_{\beta })}$

${\displaystyle \delta _{2}={\frac {1}{2}}}$, ${\displaystyle \exists A_{2}\subset X}$ where ${\displaystyle (diam(A_{2})<\delta _{2})}$ and ${\displaystyle \neg \exists \beta (A_{2}\subset U_{\beta })}$

${\displaystyle \delta _{k}={\frac {1}{k}}}$, ${\displaystyle \exists A_{k}\subset X}$ where ${\displaystyle (diam(A_{k})<\delta _{k})}$ and ${\displaystyle \neg \exists \beta (A_{k}\subset U_{\beta })}$

For all ${\displaystyle n\in \mathbb {Z} ^{+}}$, ${\displaystyle A_{n}\neq \emptyset }$ since ${\displaystyle A_{n}\not \subset U_{\beta }}$.

It is therefore possible to generate a sequence ${\displaystyle \{x_{n}\}}$ where ${\displaystyle x_{n}\in A_{n}}$ by axiom of choice. By sequential compactness, there exists a subsequence ${\displaystyle \{x_{n_{k}}\},k\in \mathbb {Z} ^{+}}$ that converges to ${\displaystyle x_{0}\in X}$.

Using the fact that ${\displaystyle {\mathcal {A}}}$ is an open covering, ${\displaystyle \exists \alpha _{0}\in J}$ where ${\displaystyle x_{0}\in U_{\alpha _{0}}}$. As ${\displaystyle U_{\alpha _{0}}}$ is open, ${\displaystyle \exists r>0}$ such that ${\displaystyle B_{d}(x_{0},r)\subset U_{\alpha _{0}}}$. By definition of convergence, ${\displaystyle \exists L\in \mathbb {Z} ^{+}}$ such that ${\displaystyle x_{n_{p}}\in B_{d}\left(x_{0},{\frac {r}{2}}\right)}$ for all ${\displaystyle p\geq L}$.

Furthermore, ${\displaystyle \exists M\in \mathbb {Z} ^{+}}$ where ${\displaystyle \delta _{M}={\frac {1}{K}}<{\frac {r}{2}}}$. So, ${\displaystyle \forall z\in \mathbb {Z} ^{+}z\geq M\Rightarrow diam(A_{M})<{\frac {r}{2}}}$.

Finally, let ${\displaystyle q\in \mathbb {Z} ^{+}}$ such that ${\displaystyle n_{q}\geq M}$ and ${\displaystyle q\geq L}$. For all ${\displaystyle x'\in A_{n_{q}}}$, notice that:

• ${\displaystyle d(x_{n_{q}},x')\leq diam(A_{n_{q}})<{\frac {r}{2}}}$ because ${\displaystyle n_{q}\geq M}$.
• ${\displaystyle d(x_{n_{q}},x_{0})<{\frac {r}{2}}}$ because ${\displaystyle q\geq L}$ which means ${\displaystyle x_{n_{q}}\in B_{d}(x_{0},{\frac {r}{2}})}$.

By the triangle inequality, ${\displaystyle d(x_{0},x')<r}$, implying that ${\displaystyle A_{n_{q}}\subset U_{\alpha _{0}}}$ which is a contradiction.