I can say "sin(1/x) is a continuous function on (0,1] but its graph is not path connected", which is more formal, but likely not mean anything to most of the reader. In that sense, I guess I have also lied :)

It's also false. Take any pair of points on the graph of sin(1/x) using the domain (0,1] that you just gave. Then we can write these points in the form (a,sin(1/a)), (b,sin(1/b)) such that 0 < a < b without loss of generality. The map f(t)=(t,sin(1/t)) on [a,b] is a path connecting these two points. This shows the graph of sin(1/x) on (0,1] is path connected.

This same trick will work if you apply it to the graph of ANY continuous map from a connected subset of R into R. This is what my graph example was getting at.

The "topologists sine curve" example you see in pointset topology as an example of connected but not path connected space involves taking the graph you just gave and including points from its closure as well.

Think about the closure of your sin(1/x) graph here. As you travel towards the origin along the topologists sine curve graph you get arbitrarily close to each point along the y-axis between -1 and 1 infinitely often. Why? Take a horizontal line thru any such point and look at the intersections between your horizontal line and your y=sin(1/x) curve. You can make a limit point argument from this fact that the closure of sin(1/x)'s graph is the graph of sin(1/x) unioned with the portion of the y-axis from -1 to 1 (inclusive).

Path connectedness fails because there is no path from any one of the closure points you just added to the rest of the curve (for example between the origin and the far right endpoint of the curve).

A better explanation of the details here would be in the connectedness/compactness chapter in Munkres Topology textbook it is example 7 in ch 3 sec 24 pg 157 in my copy.

However, I like to push back on the assumption that, in the context of teaching continuous function, the underlying space needs to be bounded: one of the first continuous function student would encounter is the identity function on real, which has both a infinite domain and range.

This is fine. I stated boundedness as an additional assumption one might require for pragmatic reasons. It's not mandatory. But it's easy to imagine somebody trying to be clever and pointing out that if we allow the domain or range to be unbounded we still have problems. For example you literally cannot draw the identity function in full. The identity map extends infinitely along y=x in both directions. You don't have the paper, drawing utensils or lifespan required to actually draw this.