The inclusions in the language-class ladder are proper

If a language is regular, choose a DFA $D$ recognizing it. Make one CFG whose variables are the states of $D$, whose start variable is the start state, whose productions are $q\to a\delta (q,a)$ for each transition, and whose accepting states also have productions $q\to \epsilon$. Then the grammar generates exactly the words accepted by $D$. Hence every regular language is context-free.

Every decidable language is recognizable by running its decider and accepting exactly when the decider accepts. Thus $\text{decidable}\subseteq \text{computably enumerable}\mathrm{.}$

The language $L_{ab}:=\{a^n{b}^n:n\ge 0\}$ is context-free, for example via the grammar $S\to aSb\mid \epsilon$. It is not regular: if it were regular, [L1] would give a pumping length $p$, but $a^p{b}^p$ cannot be pumped inside the first $p$ symbols without changing the number of $a$'s and leaving the number of $b$'s fixed. Therefore $\text{regular}⊊\text{context-free}\mathrm{.}$

Relabel the binary alphabet of $A_{TM}$ injectively by the two chosen symbols $a,b\in \mathrm{\Sigma }$. This computable letter-by-letter bijection preserves recognizability and decidability. By [L5] and [L6], the relabeled language over $\mathrm{\Sigma }$ is computably enumerable but not decidable. Therefore $\text{decidable}⊊\text{computably enumerable}\mathrm{.}$

Every context-free language is decidable by [L2], so $\text{context-free}\subseteq \text{decidable}\mathrm{.}$

The language $L_{aba}:=\{a^n{b}^n{a}^n:n\ge 0\}$ is decidable by checking the three blocks and comparing their lengths. It is not context-free: if $p$ were a pumping length from [L3], then a pumping window of length at most $p$ in $a^p{b}^p{a}^p$ could meet at most two consecutive blocks and could not meet both $a$-blocks. Pumping a nonempty part therefore changes at least one of the three required counts without changing all three equally, or destroys the block order. This contradicts [L3]. Hence $\text{context-free}⊊\text{decidable}\mathrm{.}$

Steps 2.1, 4.1, and 2.2 prove that all three containments are strict over the fixed alphabet $\mathrm{\Sigma }$.