I am assuming that you already got the solution from this: https://mathoverflow.net/questions/413632/special-version-of-tonelli-s-theorem.

However, the solution has some mistakes and I corrected some information in the link to get the full solution as I see it:

Assume that $b>a$, and we can argue as follows: since $f$ is bounded then there is a real number $m \in \mathbb R$ such that

$$m\le f(x,\zeta), \quad \forall (x,\zeta) \in [a,b] \times \mathbb{R}.$$

From the properties of $g$ we get

$$m (b-a) + C \int_a^b |u'(x)|^r dx \leq J[u] \Rightarrow m+ C \| u'\|_{L^r[a,b]}^r \leq J[u]\,\,\, \forall u \in X.$$

We can see that $J[u]$ is bounded below and from the definition of the infimum there is a minimizing sequence $\{u_n\}_{n\in \mathbb{N}} \subset X$ such that

$$\underset{n \to \infty}{\lim} J[u_n] = \inf \{ J[u] | u \in X \}> -\infty \,\, \text{ in } \mathbb{R},$$

and hence, $\{ u_n'\}_{n \in \mathbb{N}}$ is uniformly bounded. W.l.o.g assume that $J[u_n]$ is decreasing, then there is $N>0$ such that $\forall n > N$ we have
$$\| u'_n\|_{L^r[a,b]} \leq \left(\frac{J[u_N] -m}{C} \right)^\frac{1}{r}.$$

Now, since $\{u_n\}$ is uniformly bounded and equicontinuous, that is from the uniform boundedness of $\{u_n'\}$ there are $M>0$ such that $|u_n'| < M $ $\forall n$. Let $\epsilon >0$ and $n \in \mathbb N$ and for all $x,y \in [a,b]$ such that $|x-y| < \frac{\epsilon}{M^2 (b-a)}$ we have
\begin{align*}
| u_n(x) -u_n(y)|&=\left|\int_x^y u_n'(t) dt \right|\\
&\le \sqrt{y-x} \left( \int_x^y |u_n'(t)|^2 dt\right)^\frac{1}{2}\\
&\le M^2 (y-x)\\
&<\epsilon.
\end{align*}
then according to Arzela-Ascoli theorem there is a subsequence $\{ u_{n_k} \}_{k \in \mathbb{N}}$ and $\overline{u} \in AC[a,b]$ such that $u_{n_k} \to \overline{u}$ uniformly, and $u'_{n_k} \to \overline{u}'$. W.l.o.g. we may assume that $u_n\to \bar u$ uniformly. So, since $L^r$ is reflexive, $1 < r < \infty$, by the Banach–Alaoglu theorem, passing again to a subsequence, w.l.o.g. we may assume that $u_n'\to v$ for some $v\in L^r$, where $r':=r/(r-1)$. Now, by Mazur's lemma, for each natural $n$ there exist a natural $N_n\ge n$ and nonnegative real numbers $a_{n,k}$ for $k\in\{n,\dots,N_n\}$ such that $\sum_{k=n}^{N_n}a_{n,k}=1$ and
\begin{equation*}
v_n:=\sum_{k=n}^{N_n}a_{n,k} u_k'\to v \tag{0}
\end{equation*}
in $L^r$.
For $x\in[a,b]$, let now
\begin{equation*}
w_n(x):=u_n(a)+\int_a^x v_n(t)\,dt
=u_n(a)-\sum_{k=n}^{N_n}a_{n,k}u_k(a)+\sum_{k=n}^{N_n}a_{n,k}u_k(x). \tag{1}
\end{equation*}

Since $u_n\to \bar u$ uniformly and the $u_n$'s are uniformly bounded, we see that $w_n\to \bar u$ uniformly and the $w_n$'s are uniformly bounded. Therefore and because $f$ is continuous, we have
\begin{equation*}
J_1[w_n] := \int_a^b f(x,w_n(x))\, dx\to J_1[\bar u]=\lim_{n\to \infty} J_1[u_n].
\end{equation*}
Also, by the convexity of $g(x,\xi)$ in $\xi$,
\begin{equation*}
J_2[w_n] := \int_a^b g(x,w_n'(x))\, dx
\le\sum_{k=n}^{N_n}a_{n,k}J_2[u_k].
\end{equation*}
And, $J[w_n]=J_1[w_n]+J_2[w_n]$. So,
\begin{equation*}
\begin{aligned}
\limsup_{n\to \infty} J[w_n]&\le \lim_{n\to \infty} J_1[w_n]+\limsup_{n\to \infty} J_2[w_n] \\
&\le \lim_{n\to \infty} J_1[u_n]+\sum_{k=n}^{N_n}a_{n,k}\limsup_{n\to \infty} J_2[u_n] \\
&\le \lim_{n\to \infty} J_1[u_n]+\limsup_n J_2[u_n] \\
&= \limsup_{n\to \infty} (J_1[u_n]+J_2[u_n]) \\
&= \limsup_{n\to \infty} J[u_n]= \lim_{n\to \infty} J[u_n]\\
&=\inf_{u\in X} J[u].
\end{aligned}
\end{equation*}
So, passing to a subsequence, w.l.o.g. we may assume that
\begin{equation*}
J[w_n]\to\inf_{u\in X} J[u].
\end{equation*}

Recall that $w_n\to \bar u$ uniformly. So, in view of (1) and (0),
\begin{equation*}
\bar u(x)=\bar u(a)+\int_a^x v(t)\,dt
\end{equation*}
for $x\in[a,b]$, so that $\bar u\in \operatorname{AC}([a,b])$ and $\bar u'=v$ almost everywhere (a.e.).
It also follows that $w_n'=v_n\to v=\bar u'$ in $L^r$ and hence in measure. So, by the continuity of $f$ and $g$ and the Fatou's lemma,
\begin{equation*}
J[\bar u]=J[\lim_n w_n]\le\liminf_n J[w_n]=\lim_n J[w_n]=\inf_{u\in X} J[u].
\end{equation*}
From the uniform convergence of $u_n \to \overline{u}$ we can see that $\overline{u}(a) = \alpha$ and $\overline{u}(b)=\beta$. Thus, $\bar u$ is a minimizer of $J[u]$ over $u\in X$.