The most common definitions of continuity agree on the fact that a function can be continuous only on points of its domain.
Asking whether $f(x)=1/x$ is continuous is like asking what's the preferred food of unicorns.
You're being misled by the phrase "point of discontinuity". Well, the truth is that a continuous function can many points of discontinuity. It's just an unfortunate terminology that I find being an endless source of misunderstandings. The terminology is due to an old fashioned way of thinking to continuity: it marks a “break” in the graph. However, the concept that a function is continuous if “it can be drawn without lifting the pencil” is a wrong way to think to continuity. The function
$$
f(x)=
\begin{cases}
0 & \text{if $x=0$,}\\
x\sin(1/x) & \text{if $x\ne0$}
\end{cases}
$$
is everywhere continuous, but nobody can really think to draw its graph without lifting the pencil. Can you?
The fact that $1/x$ (defined on the real line except $0$) has a point of discontinuity doesn't mean that the function is not continuous somewhere. Indeed it is continuous at each point of its domain.
Prompted by a comment, I'll add that a function can be defined at a point an not be continuous at it. The easiest example is the Dirichlet function
$$
D(x)=
\begin{cases}
0 & \text{if $x$ is irrational,}\\
1 & \text{if $x$ is rational}
\end{cases}
$$
which is continuous nowhere.
So a function can certainly be noncontinuous (I purposely avoid discontinuous) at a point where it is defined.
Returning to the function $f(x)=1/x$, one can specify any subset of the real numbers as its domain, so long as it doesn't contain $0$. When no domain is explicitly specified, it's customary to use the largest subset of the reals where the expression makes sense, in this case it is $\mathbb{R}\setminus\{0\}$.
It's surely possible to define a function $g$ that extends $f$ in $0$; the function $g$ cannot, however, be continuous, because the limit of $g$ at $0$ can't be the value $g(0)$.
OK, next thought - the function $f(\frac pq)=\frac1q$ and zero elsewhere. That's closer; it has limits of zero everywhere. But then each rational is a removable discontinuity, not a jump discontinuity. Closer, but still not it.
The next idea after that: let's build an increasing function with jumps at every rational. Let $g$ be an enumeration of the rationals; for each rational $r$, $g(r)$ is a different positive integer $n$. Then, define
$$f(x) = \sum_{r\in\mathbb{Q},r\le x}\frac1{g(r)^2+g(r)}$$
Since $\sum_n \frac1{n^2+n}$ converges (to $1$), that sum is finite for every $x$.
Choose some arbitrary $x$ and $\epsilon>0$. Let $n$ be such that $\epsilon\ge\frac1n$. There are only finitely many values $r_1,r_2,\dots,r_n$ with $g(r_i)\le n$. If we choose $\delta$ such that $(x,x+\delta)$ contains none of these $r_i$, then for $y\in (x,x+\delta)$,
$$f(y)-f(x)=\sum_{r\in\mathbb{Q},x<r\le y}\frac1{g(r)^2+g(r)} \le \sum_{j=n+1}^{\infty}\frac1{j^2+j}=\frac1{n+1}<\epsilon$$
From that, $\lim_{y\to x^+}f(y)=f(x)$ for all $x$. We have limits from the right.
For limits from the left, consider the variant function
$$f^*(x)=\sum_{r\in\mathbb{Q},r< x}\frac1{g(r)^2+g(r)}$$
This $f^*$ is equal to $f$ except at the rationals, where $f(r)-f^*(r)=\frac1{g(r)^2+g(r)}$. Again, choose arbitrary $x$ and $\epsilon>0$, and let $n$ be such that $\epsilon\ge \frac1n$. Find $\delta$ such that $(x-\delta,x)$ contains none of the $n$ points $r_i$ with $g(r_i)\le n$. Then, for $y\in (x-\delta,x)$,
$$f^*(x)-f(y) = \sum_{r\in\mathbb{Q},y\le r< x}\frac1{g(r)^2+g(r)} \le \sum_{j=n+1}^{\infty}\frac1{j^2+j}=\frac1{n+1}<\epsilon$$
From that, $\lim_{y\to x^-}f(y) = f^*(x)$ for all $x$, and we have limits from the left.
Of course, these limits $\lim_{y\to x^+}f(y)=f(x)$ and $\lim_{y\to x^-}f(y) = f^*(x)$ differ for every rational $x$, so there's a jump discontinuity at every rational.
With $f$ discontinuous at a dense set of points, it fails to be continuous on any interval, and can't be a piecewise continuous function. Done. We have our example.
I defined this as a function from $\mathbb{R}$ to $\mathbb{R}$, but it's easy to get a function on a smaller interval. Restricting $f$ works, as does using an enumeration of the rationals in that smaller interval.
Best Answer
A very easy way to construct a function that is piecewise without being "obviously piecewise" is functions defined in terms of limits:
$$f(x) = \lim_{a \to +\infty} \exp\left(-ax^2\right) = \begin{cases}1, & x = 0 \\ 0, & \text{otherwise}\end{cases}$$
This example has the advantage of being easily-comprehensible to beginning calculus students.