Piecewise Continuously Differentiable Function/Definition 2 is Continuous

Theorem
Let $f$ be a real function defined on a closed interval $\left[{a \,.\,.\, b}\right]$.

Let $f$ satisfy Piecewise Continuously Differentiable Function/Definition 2.

Then $f$ is continuous.

Proof
Since $f$ satisfies Piecewise Continuously Differentiable Function/Definition 2, there exists a finite subdivision $\left\{{x_0, \ldots, x_n}\right\}$ of $\left[{a \,.\,.\, b}\right]$, $x_0 = a$ and $x_n = b$, such that $f$ is continuously differentiable on $\left[{x_{i−1} \,.\,.\, x_i}\right]$, the derivatives at $x_{i−1}$ and $x_i$ understood as one-sided derivatives, for every $i \in \left\{{1, \ldots, n}\right\}$.

By Differentiable Function is Continuous and $f$ being differentiable on $\left[{x_{i−1} \,.\,.\, x_i}\right]$, $f$ is continuous at every point of $\left[{x_{i−1} \,.\,.\, x_i}\right]$, the continuities at $x_{i−1}$ and $x_i$ being one-sided, for every $i \in \left\{{1, \ldots, n}\right\}$.

We use this result to go through every point of every $\left[{x_{i−1} \,.\,.\, x_i}\right]$ in order to establish the continuity of $f$ there:

$f$ is continuous at every point of every open interval $\left({x_{i−1} \,.\,.\, x_i}\right)$ for every $i \in \left\{{1, \ldots, n}\right\}$ because $\left({x_{i−1} \,.\,.\, x_i}\right)$ is a subset of $\left[{x_{i−1} \,.\,.\, x_i}\right]$ where $f$ is continuous.

$f$ is continuous at $x_0$ because $f$ is right-continuous there and $x_0$ (=$a$) is the leftmost point in the domain of $f$.

$f$ is continuous at $x_n$ because $f$ is left-continuous there and $x_n$ (=$b$) is the rightmost point in the domain of $f$.

$f$ is continuous at every point $x_i$, $i \in \left\{{1, \ldots, n-1}\right\}$, because $f$ is both left- and right-continuous at these points.

All in all, we have found that $f$ is continuous at every point of $\left[{x_{i−1} \,.\,.\, x_i}\right]$ for every $i \in \left\{{1, \ldots, n}\right\}$.

$f$ is therefore continuous at every point of the union of those intervals: $\displaystyle \bigcup_{i \mathop = 1}^n \left[{x_{i−1} \,.\,.\, x_i}\right]$.

Since $\displaystyle \bigcup_{i \mathop = 1}^n \left[{x_{i−1} \,.\,.\, x_i}\right]$ equals $\left[{a \,.\,.\, b}\right]$, the domain of $f$, we conclude that $f$ is continuous.