In a (classical) lagrangian field theory, the configuration space $\mathcal C$ of the system is a space of field configurations. A field configuration (or just "field" for short) is usually taken to be a function $\phi:\mathcal M\to T$ where $M$ is a manifold and $T$ is some set, often a manifold or vector space or both, called the target space of the field. The configuration space $\mathcal C$ is then taken to be some sufficiently smooth (when a notion of smoothness can be defined) subset of the set of all possible fields. The lagrangian is then a function $L:\mathcal C\to\mathcal C$;
\begin{align}
\phi\mapsto L[\phi]
\end{align}
Namely, the lagrangian maps a particular field configuration, to another field configuration. Often, one considers a field theory for which the lagrangian can be written as local local density, but this is not strictly speaking necessary. The action of the theory can then be defined as the integral of $L[\phi]$ over $M$;
\begin{align}
S[\phi] = \int_M \, d^Dx\,L[\phi](x).
\end{align}
Note. My terminology and notation here are a bit non-standard in some contexts. For example, in relativistic physics (field theory) the Lagrangian will usually map a field configuration $\phi$ to a function $L[\phi]$ of time, and then this function of time will be integrated to yield the action. It's not hard in practice to translate between conventions.
One can then define what it means for the action to possess symmetry. In particular, given a mapping $F:\mathcal C\to \mathcal C$ of the manifold on which field configurations are defined to itself, one says that the action is invariant under $F$ provided
\begin{align}
S[F(\phi)] = S[\phi]
\end{align}
for all $\phi\in\mathcal C$. For "continuous" transformations, one can also define notions of symmetry that don't involve full invariance, but let's keep the discussion simple at this point.
Example. A common toy theory considered as the first example in most relativistic field theory texts is that of a single, free, real Lorentz scalar defined on Minkowski space (I'll use metric signature $+ - - -$). In this case, we have
\begin{align}
M = \mathbb R^{3,1}, \qquad T = \mathbb R
\end{align}
and $\mathcal C$ is a space of sufficiently smooth functions $\phi:\mathbb R^{3,1}\to\mathbb R$ that satisfy certain desired boundary conditions. The Lagrangian of such a theory is
\begin{align}
L[\phi](x) = \mathscr L(\phi(x), \partial_0\phi(x), \dots \partial_3\phi(x))
\end{align}
where $\mathscr L$ is the Lagrangian density defined by
\begin{align}
\mathscr L(\phi, \partial_0\phi, \dots, \partial_3\phi)
&= \frac{1}{2}\partial_\mu\phi\partial^\mu\phi - \frac{1}{2}m^2\phi^2.
\end{align}
Given any Lorentz transformation $\Lambda$, one can define a transformation $F_\Lambda:\mathcal C\to\mathcal C$ as follows:
\begin{align}
F_\Lambda(\phi)(x) = \phi(\Lambda^{-1} x).
\end{align}
A short computation then shows that the Lagrangian is a Lorentz scalar under this transformation, namely
\begin{align}
L[F_\Lambda(\phi)](x) = L[\phi](\Lambda^{-1}x).
\end{align}
In fact, this is essentially done for you on page 36 of Peskin. It follows from this that the action is invariant under $F$;
\begin{align}
S[F_\Lambda(\phi)] = \int_{\mathbb R^{3,1}} d^4x\, L[\phi](\Lambda^{-1}x) = \int_{\mathbb R^{3,1}} d^4x\, L[\phi](x) = S[\phi].
\end{align}
since the measure $d^4 x$ is Lorentz-invariant. Notice, in particular, that the fact that the Lagrangian transformed as a Lorentz scalar (namely precise in the same way as the scalar field $\phi$ was defined to transform), immediately led to invariance of the action.
Furthermore, suppose that $\phi$ is a field configuration that leads to stationary action, then we can also show that a Lorentz-transformed field leads to a stationary action using Lorentz invariance of the action. To see this, recall that the variational derivative in the direction of a field configuration $\eta$ is defined as follows:
\begin{align}
\delta_\eta S[\phi] = \frac{d}{d\epsilon}S[\phi+\epsilon\eta]\Big|_{\epsilon=0}
\end{align}
Now, suppose that $\phi$ is a stationary point of the action, namely that $\delta_\eta S[\phi] = 0$ for all admissible $\eta$, then for all such $\eta$ we have
\begin{align}
\delta_{F_\Lambda(\eta)} S[F_\Lambda(\phi)]
&= \frac{d}{d\epsilon} S[F_\Lambda(\phi) + \epsilon F_\Lambda(\eta)]\Big|_{\epsilon = 0} \\
&= \frac{d}{d\epsilon} S[F_\Lambda(\phi+\epsilon_\eta)]\Big|_{\epsilon = 0} \\
&= \frac{d}{d\epsilon} S[\phi+\epsilon_\eta]\Big|_{\epsilon = 0} \\
&= 0
\end{align}
Now set $\eta = F_\Lambda^{-1}(\xi)$, then the computation we just performed shows that
\begin{align}
\delta_{\xi} S[F_\Lambda(\phi)] =0.
\end{align}
for all admissible field configurations $\xi$. In other words, the Lorentz transformed scalar is also a stationary point of the action. Notice that this demonstration holds for any Lorentz transformation, not just boosts.
Addendum.
As pointed out in the comments, the argument at the end about variational derivatives hinges on linearity of $F_\Lambda$. This can be demonstrated as follows:
\begin{align}
F_\Lambda(a\phi+b\psi)(x)
&= (a\phi+b\psi)(\Lambda^{-1}x) \\
&= a\phi(\Lambda^{-1}x) + b\psi(\Lambda^{-1}x) \\
&= aF_\Lambda(\phi)(x) + bF_\Lambda(\psi)(x).
\end{align}
Let me make some remarks about the mapping $F:\mathcal C\to \mathcal C$; a symmetry of the action. If there exists a mapping $f_T:T\to T$ on the target space that induces this mapping, namely
\begin{align}
F(\phi)(x) = f_T(\phi(x)),
\end{align}
then $F$ is called an internal symmetry. On the other hand, if there is a mapping $f_M:M\to M$ on the base manifold $M$ that induces this mapping, namely
\begin{align}
F(\phi)(x) = \phi(f_M(x)),
\end{align}
then $F$ is called a base manifold symmetry (or more commonly a spacetime symmetry since in the context of relativistic field theory, the base manifold is a spacetime like Minkowski space.)
Furthermore, the mapping $F:\mathcal C \to\mathcal C$ on the field configuration space is often, as in the scalar field example, a group action of some group $G$ on $\mathcal C$. This means that to each $g\in G$, we associate a mapping $F_g:\mathcal C\to \mathcal C$ such that the mapping $g\mapsto F_g$ is a homomorphism of the group $G$. In practice, the group $G$ is sometimes a group of symmetries that naturally acts on the base manifold, and sometimes $G$ a group of symmetries that naturally acts on the target space (or even both when the $M=T$). In any event, this group action is usually obtained by composing a target space group action $(f_T)_g:T\to T$ with a base manifold group action $(f_M)_g:M\to M$. More explicitly, for each $g\in G$, we can define mappings $(F_T)_g:\mathcal C\to \mathcal C$ and $(F_M)_g:\mathcal C\to\mathcal C$ as follows:
\begin{align}
(F_T)_g(\phi)(x) = (f_T)_g(\phi(x)), \qquad (F_M)_g(\phi)(x) = \phi((f_M)_g(x))
\end{align}
and then the full group action $F_g:\mathcal C\to\mathcal C$ is defined by the composition of these two;
\begin{align}
F_g = (F_T)_g\circ (F_M)_g
\end{align}
or more explicitly
\begin{align}
F_g(\phi)(x) = (f_T)_g(\phi((f_M)_g(x))).
\end{align}
Now, this is all a bit abstract, so let's write out what all of these objects would be for the scalar field example:
\begin{align}
G &= \mathrm{SO}(3,1) \\
g &= \Lambda\\
(f_T)_g(\phi(x)) &= \phi(x) \\
(f_M)_g(x) &= \Lambda^{-1}x \\
(F_T)_g(\phi)(x) &= \phi(x) \\
(F_M)_g(\phi)(x) &= \phi(\Lambda^{-1}x) \\
F_g(\phi)(x) &= \phi(\Lambda^{-1}x)
\end{align}
Notice that $f_T$ is simply the identity mapping on the target space. This is precisely what we mean when we say that the scalar field is a Lorentz scalar. On the other hand, for a Lorentz vector, the target space itself would be Minkowski space $\mathbb R^{3,1}$, and the target space group action would be
\begin{align}
(f_T)_\Lambda(A(x)) = \Lambda A(x),
\end{align}
namely, there is an internal symmetry in which the vector indices on the field transform non-trivially. In components, which is how you'll see this written in Peskin for example, the right hand side would be written as $\Lambda^\mu_{\phantom\mu\nu} A^\mu(x)$.
Reparametrization invariance of the worldline simply means that the physics of a problem cannot depend upon the particular choice of parameter a physicist makes, the pramaeter $\tau$ per se has no meaning, you can still choose to parametrize the wordline $x^\mu(\tau)$ with something that has a meaning like the proper time, which is only a convenient choice.
You can say the invariance is a mathematical consequence of the fact that you don't want to introduce in the theory a preferred time frame (i.e. you don't want to fix a frame of reference from the start) and that creates a redundacy: (an additional non physical symmetry, a gauge symmetry) the freedom to fix the parameter $\tau$. If for exemple you choose proper time, you would have $\tau=\tau(x^0,v)$ where $x^0$ is your coordinate time and $v$ is the velocity of the particle relative to you. This means of the $D$ coordinates you had, only $D-1$ are indipendent. In a more sophisticated language that means the Hamiltonian of the system is of pure constraint, i suggest you to read Dirac's "Lectures on quantum mechanics" if you want to better understand lagrangian and hamiltonian mechianics in the context of special relativity.
As a concluding remark on this issue, i will say that parametrization invariance means that once you compute relations between observables quantities the dependence on the particular paramter you chose must vanish, if it wasn't so, differnt observers would see different physics just by having fixed $\tau$ in a different way.
To your practical problems:
You have written $\mathcal{L}=-m\sqrt{\dot{X}^\mu\dot{X}_\mu}$ which makes for an invariant action under reparametrization. In classical mechanics in the context of special relativity the only known four force that has all the right proprieties (Lorentz+rep, you can easily check this) is the electromagnetic one $$\tag 1\mathcal{L}_I=e \dot{X}_\mu A^\mu$$ where $A^\mu$ is the usual E.M. four potential. You can think of this as: you can treat (without going insane) in a fully lorentz covariant way only the foundamental interactions of nature. If you want to study the potential $V=\frac{1}{2}kx^2$ which has a preferred frame, or others non relativistic potentials, you will have to fix from the start the frame of reference and $\tau$ and use and had oc lagrangian which will not have any obvious lorentz symmetry, for example: $$\tag 2\mathcal{L}_{adhoc}=-mc^2\sqrt{1-\beta^2}-V(x)$$ with this you can solve most practical problems. This doesn't mean that the theory with $\mathcal{L}_{adhoc}$ isn't consistent, it just means that it's valid for one particular choice of the frame of reference. If you want to know the soultion in another frame of reference, or reparametrize it, you will have to start back from scratch and define a new lagrangian for the new inertial frame.
Best Answer
joshphysics's answer was excellent, and I'd like to draw attention to another aspect of the question.
Your action,
$S=-m\int\sqrt{1-u^{2}}dt$
Can be written:
$S=-m\int\sqrt{(dt)^{2}-(udt)^{2}}$
$=-m\int\sqrt{(dt)^{2}-(dx)^{2}} = -m\int ds$
And the transformed action $S' = -m \int ds'$.
So, for the action to be Lorentz invariant, it suffices to show that the interval, $s$, is Lorentz invariant, i.e. $s=s'$:
Say $s^2 = t^2 - x^2$; $x' = \gamma (x-vt)$; $t' = \gamma (t-vx)$; and finally $\gamma = (1-v^2)^{-1/2}$. Then,
$(s')^2 = (t')^2 - (x')^2$
$=\gamma^2((t-vx)^2 - (x-vt)^2) = \gamma^2(t^2 +v^2x^2 - x^2+ -v^2t^2)$
(Cross terms cancel)
$= (1-v^2)^{-1}(t^2(1-v^2) -x^2(1-v^2)) = t^2-x^2$
So $ds'=ds$, and consequently your action is Lorentz invariant.