In the worked solutions from my school textbook they present the following to show that the spheres $\left|\;\underset{\sim}{r}-\begin{bmatrix} 5 \\-6\\3\\\end{bmatrix}\;\right|=7$ and $\left|\;\underset{\sim}{r_1}-\begin{bmatrix} -3 \\2\\7\\\end{bmatrix}\;\right|=5$ only have a single point of contact.
The cartesian equations for $\underset{\sim}{r}$ and $\underset{\sim}{r_1}$ are $(x-5)^2+(y+6)^2+(z-3)^2=49$ and $(x+3)^2+(y-2)^2+(z-7)^2=25$ respectively. This means that the cartesian equations in terms of $x$ will be $(x-5)^2=7^2$ and $(x+3)=5^2$.
Equating these two equations to find the intersecting point on the $x$-axis $(x-5)^2-49=(x+3)^2-25$, which gives $x=\frac{1}{2}$. As there is only a single value for the two spheres to intersect on the $x$-axis, the spheres touch each other at a single point.
Now I don't understand how this shows the required statement in two ways. One, another contact point could surely have the same $x$ coordinate but just a different $z$ or $y$ coordinate. But even then, if we repeat the process for $y$ and $z$, doesn't this way show that any two spheres touch at only one point because equating the two quadratics will always cancel out the squared term, leaving one solution? So could someone explain if I'm right to say this proof doesn't work, and if so, could they provide an actual proof as to whether two spheres only have one point of contact?
Yes if that is the exact given solution, then it is wrong. Also, $(x+3)^2+(y-2)^2+(z-7)^2=25$ does not even intersect x-axis. You can check so by plugging in $y = z = 0$.
In general, if we have two spheres with centers $O_1$ and $O_2$, and radii $r_1$ and $r_2$ respectively, then
$i) ~ |O_1O_2| \gt r_1 + r_2 ~$ means both spheres do not intersect.
$ii) ~ |O_1O_2| = r_1 + r_2 ~$ means both spheres are tangent to each other i.e. they have single point of contact.
$iii) ~ |O_1O_2| \lt r_1 + r_2 ~$ means both spheres intersect each other (in a circle).
In this case, we see that the distance between their centers $|O_1 O_2|$ is equal to sum of their radii.
$|O_1O_2| = \sqrt{(3 + 5)^2 + (-2-6)^2 + (-7+3)^2} = 12 = 7 + 5$
So the spheres must be tangent to each other.