Why is Shamir Secret Sharing not secure against active adversaries out-of-the-box?
Here is an active attack on the privacy of out-of-the-box SSS.  For this attack, we'll assume that the attacker (without a valid share) is allowed to participate (with $T-1$ friends with honest key shares), jointly use the protocol to recover a 'shared secret' (which might not be the real shared secret); we'll assume that this shared secret recovery process is done by some MPC-secure method so that the attacker learns nothing other than the final value.  We also assume that everyone's $x$ coordinate is public.

First off, the attacker picks a value $x_a$ which is an $x$ coordinate which is not the $x$-coordinate shared by anyone else.  If the dealer were to generate a share with that $x_a$ value, it would generate a $y$-coordinate that we'll call $y_a$; obviously, the attacker does not know this $y_a$ value.

Then, the attacker then gets a set of $T-1$ friends, and have them recover a shared secret value using their shares (and the attacker will contribute the value $(x_a, 0)$; (actually, a nonzero value for the $y$ coordinate would work, but make the explanation slightly more involved); they'll go through the SSS recovery process, and obtain a value which we'll denote as $S_0$.

If we call the real shared secret $S$, then what the attacker can do is derive a value $T_0$ (which depends on the $x$-coordinates of him and his friends), and derive the linear equation:

$$S = T_0 y_a + S_0$$

This is a linear equation in two unknowns ($S, y_a$), and $T_0 \ne 0$, and so the attacker cannot derive any information about $s$ from it.  This can also be derived from the security properties of SSS; the attacker and his $T-1$ friends only have $T-1$ valid shares, and so of course they cannot recover $S$.

What the attacker then does is get together a different set of $T-1$ friends (there can be some overlap, but there must be at least one difference), and do the same thing again (with the attacker using the same $(x_a, 0)$ share.  They do the same thing, deriving a 'shared secret' value $S_1$, and the attacker ends up with a second linear equation:

$$S = T_1 y_a + S_1$$

Now, if $T_0 \ne T_1$ (which is likely if the two sets of friends are distinct), the attacker has two linear equations in two unknowns; it's trivial to solve for $S$.