Show that, if every tangent line of a unit speed curve $\beta$ passes through a fixed point $p$, then $β$ is a line. Just wondering whether my proof is valid or not, feedback would be very appreciated! 
If the tangent line $T$ always intersects a particular point $p$ then it must be the case that $T$ is constantly pointing either directly towards $p$ or directly away from $p$. We assume that $\beta$ is smooth and hence we are forced to pick one of the two options ($T$ can't instantaneously switch directions if $\beta$ is smooth.) Without loss of generality, assume $T$ is pointing towards $p$.
$T$ is a vector composed of magnitude and direction. Since $T$ is a unit vector, its magnitude is never changing. Since $T$ is always pointing directly towards $p$, its direction is never changing. Hence, $T$ is a constant vector. Then,
$T$ is constant $\Rightarrow |T'| = \kappa = 0$. Given that $\kappa = 0$, $\beta$ must be a straight line. 
 A: First, a little feedback:
"Since $T$ is always pointing directly towards $p$, its direction is never changing."
I don't quite see how this follows from our OP talrefae's argument; I think a little more work is needed to establish this.
These things being written, I believe the following argument contains a correct demonstration that $T(s)$ is, in fact, constant, and hence that it's direction does not change:
I'm going to assume $p$ does not lie on the curve $\beta(s)$.
First of all, what is the line $l$ tangent to $\beta(s)$ for some particular value of $s$?  Well, one way to express it is in the form
$l(\tau, s) = \beta(s) + \tau T(s); \tag 1$
that is, for some $s$, the "curve" $l(\tau, s)$, considered as a function of the parameter $\tau$, takes the value $\beta(s)$ when $\tau = 0$, and it's tangent vector $T(s)$ is constant with respect to $\tau$:  it is well-known, and easy to see, that for fixed $s$, $l(\tau, s)$ is the line through $\beta(s)$ in the direction of $T(s)$.  Since by definition
$\dot \beta(s) = T(s), \tag 2$
we see that (1) indeed presents the line tangent to the curve $\beta(s)$ at $s$; since we are given that this line passes through the point $p$, for some value of $\tau(s)$ depending on $s$, we have
$p = \beta(s) + \tau(s) T(s); \tag 3$
it is easy to see that $\tau(s)$ is a differentiable function of $s$, since (3) may be written
$\tau(s) T(s) = p - \beta(s), \tag 4$
whence, since $T(s) \cdot T(s) = 1$ (because $\beta(s)$ is a unit speed curve), 
$\tau(s) = \tau(s) T(s) \cdot T(s) = (p - \beta(s)) \cdot T(s), \tag 5$
so $\tau$ is a differentiable function of $s$.  We may thus take $d/ds$ of each side of (3):
$0 = \dot p = \dot \beta(s) + \dot \tau(s) T(s) + \tau(s) \dot T(s), \tag 6$
where $\dot p = 0$ since $p$ is fixed; we substitute (2) into this equation and find
$T(s) + \dot \tau(s) T(s) + \tau (s) \dot T(s) = 0, \tag 7$
or
$(\dot \tau(s) + 1)T(s) + \tau(s) \dot T(s) = 0; \tag 8$
now
$T(s) \cdot T(s) = 1 \Longrightarrow T(s) \cdot T'(s) = 0; \tag 9$
thus,
$\dot \tau(s) + 1 = (\dot \tau(s) + 1)T(s) \cdot T(s) + T(s) \cdot T'(s) = 0, \tag{10}$
that is,
$\dot \tau(s) = -1; \tag{11}$
we note in passing that the unique solution to this with $\tau(s_0) = \tau_0$ is
$\tau(s) = s_0 -s + \tau_0 = 0 \Longleftrightarrow s = s_0 + \tau_0; \tag{12}$
we also observe that (4) further yields
$\tau^2(s) = \tau^2(s) T(s) \cdot T(s) = (\tau(s)T(s)) \cdot (\tau(s) T(s)) = (p - \beta(s)) \cdot (p - \beta(s)) \ne 0 \tag{13}$
when $p \ne \beta(s)$, which binds since we have assumed $p$ is not on the curve $\beta(s)$; (8) then leads us to 
$\dot \tau(s) + 1 = 0, \; \tau(s) \ne 0 \Longrightarrow \dot T(s) = 0 \Longrightarrow T(s) = T_0; \tag{14}$
we may now integrate (2) with $T(s) = T_0$ and $\beta(s) = \beta(s_0)$ at $s = s_0$:
$\dot \beta(s) = T(s) = T_0 \Longrightarrow \beta(s) = T_0 (s - s_0) + \beta(s_0), \tag{15}$
the equation of a straight line.
