Let L be any non-empty language over an alphabet Σ. Show that $L^2$ ⊆ $L^3$ if and only if λ ∈ L
I'm assuming by $\lambda$ you mean the empty word and by $n_0(w)$ the length of a word.

The proof for the first part is not correct: You argue that every word in $L^2$ has length at least $2$ and every word in $L^3$ has length at least $3$. From that it does not follow that every word in $L^3$ is longer than every word $L^2$, because there could also be a word of e.g. length $3$ in $L^2$. Consider for example $L = \{a\}^\star \setminus \{ \lambda \}$, that is, the set of all non-empty $a$-sequences. You have $a \in L$ and $aa \in L$, thus $aaa = a \cdot aa \in L^2$, but also $aaa = a \cdot a \cdot a \in L^3$.

Instead, what you want to say is that there is a word of minimal length $n$ in $L^2$, and because every word in $L$ has length at least one, the shortest word in $L^3$ has length at least $n+1$.

More formally: Let $\lambda \notin L$. Thus $\forall w \in L: n_0(w) &gt; 0$. Pick $w \in L^2$ such that $n_0(w)$ is minimal. Assume $w \in L^3$. Since $L^3 = L^2 \cdot L$, it follows that $w = u \cdot v$ for $u \in L^2$ and $v \in L$. Since $n_0(v) &gt; 0$, it follows $n_0(w) &gt; n_0(u)$. But then $n_0(w)$ is not minimal. Contradiction.

For the second part: Assume $L^2 \not \subseteq L^3$ and $\lambda \in L$. Thus there is $w \in L^2$ such that $w \notin L^3$. But since $L^3 = L^2 \cdot L$, it follows $w = w \cdot \lambda \in L^3$. Contradiction.