I'm actually unsure of the question itself. Not sure what it is asking for, but when I graphed it, I see a parabola with increasing $x,y,z$ values as t increases.
How can one describe the continuity of the curve? It is continuous for all values of T, would that be correct?
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If you just look at the $x$ and $y$ values (ignoring $z$), you get a definite parabola, vertical in orientation, open upwards and with vertex at the point $(0,-\frac 12)$.
Your complete graph also has $y=z$, so instead of a parabola going "up" you have one at a $45°$ angle to the $xy$ and $xz$ planes. The vertex here is the point $(0,-\frac 12,-\frac 12)$.
All the coordinates are continuous functions in $t$, so the entire curve is continuous.
More could be said in a description, such as the focal point of the parabola, but this is probably enough.
You know that $$x=t \,;\, y=\frac{t^2-1}{2} \,;\, z=\frac{t^2-1}{2}$$
Now we can change the equations without changing the curve (i.e. each step can be tracked backwards to our original description) the following way : $$x=t \,;\, y=\frac{t^2-1}{2} \,;\, z=y \\ x=t \,;\, y=\frac{x^2-1}{2} \,;\, z=y \\ y=\frac{x^2-1}{2} \,;\, z=y \\ 2y=x^2-1 \,;\, z=y \\ $$
Now, $2y=x^2-1$ is a parabola in plane, and hence a parabolic sheet in space (i.e. a vertical plane curbed by this parabola). $z=y$ is a plane. Our curve is the intersection between those two surfaces.