Proof of Continuous Real-Valued Functions on a Compact Interval are Bounded

Suppose, for contradiction, that no real number $C\ge0$ bounds $|g|$ on $[a,b]$. Then for every natural number $n$ there is $t_n\in[a,b]$ with $|g(t_n)|>n$. The sequence $(t_n)_{n\in\mathbb{N}}$ is a bounded sequence of real numbers, since $a\le t_n\le b$ for every $n$. By the Bolzano-Weierstrass theorem there are a subsequence $(t_{n_k})_{k\in\mathbb{N}}$ and a real number $t^*$ such that $t_{n_k}\to t^*$ in the sense of convergence of real sequences.

First, $t^*\in[a,b]$. Indeed, if $t^*>b$, then taking $\varepsilon=t^*-b>0$ in the definition of convergence yields some $k$ with $|t_{n_k}-t^*|<\varepsilon$, hence $t_{n_k}>t^*-\varepsilon=b$, contradicting $t_{n_k}\le b$; the case $t^*<a$ is excluded symmetrically.

Since $g$ is continuous at $t^*$ relative to $[a,b]$, there is $\delta>0$ such that $|g(t)-g(t^*)|<1$ for every $t\in[a,b]$ with $|t-t^*|<\delta$. By convergence of the subsequence, choose $k$ so large that both $|t_{n_k}-t^*|<\delta$ and $n_k>|g(t^*)|+1$ (the indices $n_k$ are strictly increasing, so $n_k\ge k$ and such $k$ exists). Then

a contradiction. Hence some $C\ge0$ satisfies $|g(t)|\le C$ for all $t\in[a,b]$.