Let $D$ be a bounded domain in $\mathbb{R}^n$ and $\Omega=D\times (0,T)$. For any $q>1$ and any $\Omega^{\prime}\Subset\Omega$, $\left\|v_k\right\|_{W_{q}^{2,1}(\Omega^\prime)<+\infty}$, where $M(\Omega^\prime)$ is independent of $k$. By the embedding theorems and a diagonalization argument, there exists a subsequence of $v_k$ (still denote it by ${v_k}$) such that $v_k\to$ some $v$ in $C^{1+\alpha,\frac{1+\alpha}{2}}_{\text{loc}}(\Omega) $ for some $0<\alpha<1$.
By the embedding theorems, if $q>n+2$, then $W_{q}^{2,1}(\Omega^{\prime})\hookrightarrow C^{1+\alpha,\frac{1+\alpha}{2}}(\overline{\Omega^\prime})$, where $\alpha=1-\frac{n+2}{q}$. Thus, $\left\| v_{k} \right\|_{C^{1+\alpha,\frac{1+\alpha}{2}}(\overline{\Omega}^{\prime})}\le M^\prime(\Omega^{\prime})$. I don't know how to use a "diagonalization argument". In the proof of Arzelà–Ascoli, we used a diagonalization argument, but Arzelà–Ascoli is for a subset of $C(M)$, where $M$ is a compact metric space. Can we use a similar argument in $C^{1+\alpha,\frac{1+\alpha}{2}}(\overline{\Omega^\prime})$?
A diagonal argument is a general technique when you want to make a proof which requires extracting a subsequence from a subsequence from a subsequence from ... from a sequence, a (countably) infinite number of times, in order to create a new sequence satisfying simultaneously many properties. You can use it in many contexts. It does not depend on the underlying space.
Say that you start from a sequence $(u^0_n)_{n \in \mathbb{N}}$, and that you know that you can extract from it a subsequence $(u^1_n)_{n \in \mathbb{N}}$ which satisfies some asymptotic property $P_1$. Now assume that you can iterate the process to extract from the sequence $(u^k_n)_{n \in \mathbb{N}}$ a subsequence $(u^{k+1}_n)_{n \in \mathbb{N}}$ satisfying the asymptotic property $P_{k+1}$. Define $$ \bar{u}_n := u^n_n \quad \text{(hence the name "diagonal")} $$ Then the sequence $(\bar{u}_n)_{n \in \mathbb{N}}$ satisifies simultaneously all the properties $P_k$ for $k \in \mathbb{N}^*$.
In the type of proof you are mentioning, the diagonal argument typically iterates over different (increasing) domains $\Omega'$ in order for the final sequence to have a property on the whole $\Omega$.
More precisely, take $\Omega_n$ an increasing sequence of open subsets $\Omega_n \Subset \Omega$. For each fixed $n$, applying your reasoning to $\Omega' = \Omega_n$, you know that $(v_k)_{k \in \mathbb{N}}$ is uniformly bounded in $W^{2,1}_q(\Omega_n)$, so also in $C^{1+\alpha,\frac{1+\alpha}{2}}(\overline{\Omega_n})$.
Now you start by extracting from $(v_k)_{k \in \mathbb{N}}$ a subsequence $(v^1_k)_{k \in \mathbb{N}}$ which converges towards some limit in $C^{1+\alpha,\frac{1+\alpha}{2}}(\overline{\Omega_1})$. From this subsequence, you extract one which converges on $\Omega_2$, and so on as detailed above.
In the end, you obtain convergence in $C^{1+\alpha,\frac{1+\alpha}{2}}_{loc}(\Omega)$ (by definition of the $C_{loc}$ space).