Does continuity always imply integrability?
Theorem: A continuous function $f: [a,b] \rightarrow \mathbb{R}$ is Riemann integrable.
Proof:
Let $f: [a,b] \rightarrow \mathbb{R}$ be a continuous function. Any function that is continuous on a compact set—such as our $f$ on $[a,b]$—is also uniformly continuous on that set$^\dagger$. This is to say, given a $\mu > 0$, we are guaranteed a $\delta > 0$ such that $|x - y| < \delta \implies |f(x) - f(y)| < \mu$ for any $x, y \in [a,b]$. Consider a partition $\mathcal{P}$ of $[a, b]$ into $n$ equal intervals of width $\displaystyle \frac{b-a}{n}$, with $n$ large enough so that $\displaystyle \frac{b-a}{n} < \delta$. Computing the difference between the upper and lower sums: \begin{align*} U(f, \mathcal{P}) - L(f, \mathcal{P}) &= \sum_{k = 1}^{n} \left(x_k - x_{k-1} \right)\Big[\operatorname{sup}\{f(x) | x \in [x_{k-1}, x_k] \} - \operatorname{inf} \{f(x) | x \in [x_{k-1}, x_k] \} \Big] \\ & \leq \left( \frac{b-a}{n} \right) \cdot n \cdot \mu \ = \ (b-a)\mu \end{align*} Given an $\varepsilon > 0$, choose $\mu$ small enough so that $\displaystyle \mu < \frac{\varepsilon}{(b-a)}$. Then $U(f, \mathcal{P}) - L(f, \mathcal{P}) < \varepsilon$, and we conclude $f$ is Riemann integrable on $[a,b]$.
$^\dagger$ See here for further discussion.
It is worthwhile to give another proof for Riemann integrability of functions which are continuous on a closed interval.
The proof below is taken from Calculus by Spivak and I must say it is novel enough. It does not make use of uniform continuity bur rather invokes mean value theorem for derivatives.
The central idea is to show that if $f:[a, b] \to\mathbb {R} $ is continuous on $[a, b] $ then the upper and lower Darboux integrals of $f$ on $[a, b] $ are equal ie $$\overline{\int} _{a} ^{b} f(x) \, dx=\underline{\int} _{a} ^{b} f(x) \, dx$$ Now to establish the above identity Spivak considers the upper Darboux integrals as a function of the upper limit of integration. Thus following Spivak we consider the function $$J(x) =\overline{\int} _{a}^{x} f(t) \, dt$$ and show that $J'(x) =f(x) $ for all $x\in[a, b] $. Similarly we have $j'(x) =f(x) $ for all $x\in[a, b] $ where $$j(x) =\underline{\int} _{a} ^{x} f(t) \, dt$$ The derivative of function $F=J-j$ vanishes everywhere on $[a, b] $ and $F(a) =0$ so that $F$ vanishes on whole of $[a, b] $.
The key point which needs to be established here is the relation $$J'(x) =f(x) =j'(x), \forall x\in[a, b] $$ and the proof is almost the same as that of first fundamental theorem of calculus. The upper Darboux integrals enjoy the same additive property as Riemann integrals and we have $$J(x+h) - J(x) =\overline{\int} _{x} ^{x+h} f(t) \, dt$$ Further given $\epsilon >0$ the continuity of $f$ at $x$ ensures the existence of a $\delta>0$ such that $$f(x) - \epsilon<f(t) <f(x) +\epsilon$$ whenever $t\in(x-\delta, x+\delta) $. If $0<h<\delta$ then the above inequality yields $$h(f(x) - \epsilon) \leq J(x+h) - J(x) \leq h(f(x) +\epsilon) $$ or $$\left|\frac{J(x+h) - J(x)} {h} - f(x) \right|\leq \epsilon$$ The same identity holds even when $-\delta<h<0$ and hence by definition of derivative we have $J'(x) =f(x) $. The proof for $j'(x) =f(x) $ is exactly the same (using lower Darboux integrals).
$f(x)=1/x$ is continuous on $[1, \infty)$, but $\int_1^{\infty} f(x) dx = \infty$.