Proof of A Continuous Function on a Closed Interval is Riemann Integrable

1. (Integrability.) The hypotheses here are exactly those of Fundamental Theorem of Calculus, Part I, on a Closed Real Interval: $a<b$, the interval $[a,b]$ is regarded as a subset of the real line, and $f:[a,b]\to\mathbb{R}$ is continuous on $[a,b]$. Clause 1 of that theorem states that for every $x\in(a,b]$ the restriction of $f$ to $[a,x]$ is Riemann integrable on $[a,x]$, where $(a,b]$ denotes $\{x\in\mathbb{R}:a<x\le b\}$ as in that theorem.

Since $a<b$ and $b\le b$, we have $b\in(a,b]$. Taking $x=b$ therefore shows that the restriction of $f$ to $[a,b]$ is Riemann integrable on $[a,b]$. That restriction is $f$ itself, because the domain of $f$ is $[a,b]$. Hence $f$ is Riemann integrable on $[a,b]$.

2. (Closed subintervals.) Let $u,v\in[a,b]$ with $u<v$. By Basic Facts about Intervals of the Real Line and Their Interior Points §closed-interval the set $[a,b]$ is an interval, so $[u,v]\subseteq[a,b]$ by Basic Facts about Intervals of the Real Line and Their Interior Points §closed-subinterval. By clause 1 of Restriction Stability of Continuity and of the Derivative, the restriction $f|_{[u,v]}$ is continuous on $[u,v]$, since $f$ is continuous on $[a,b]$ and $[u,v]\subseteq[a,b]$.

Now apply Fundamental Theorem of Calculus, Part I, on a Closed Real Interval a second time, with the real numbers $u<v$ in place of $a<b$, the closed interval $[u,v]$ regarded as a subset of the real line, and the function $f|_{[u,v]}:[u,v]\to\mathbb{R}$, which is continuous on $[u,v]$ by the previous paragraph. Its clause 1 states that for every $x\in(u,v]$ the restriction of $f|_{[u,v]}$ to $[u,x]$ is Riemann integrable on $[u,x]$. Since $u<v$ and $v\le v$, we have $v\in(u,v]$; taking $x=v$, and noting that the restriction of $f|_{[u,v]}$ to $[u,v]$ is $f|_{[u,v]}$ itself, we conclude that $f|_{[u,v]}$ is Riemann integrable on $[u,v]$.