I want to prove that $\lim_{n \to \infty} b^{1/n}$ exists for all $b ≥ 0$ and find value of such limit. There are four cases: $b \gt 1$, $b = 1$ and $1 \gt b \gt 0$ and $b = 0$. Constant cases are trivial, since we have constant sequences. I am more interested in checking my argument for non-trivial cases.
Let's take $b \gt 1$. Since $b \gt 1^n$, we have $b^{1/n} \gt 1$ by inequality of powers and roots. At the same time, since $b \gt 1^{n^2}$, we have $b^{1/n^2} \gt 1$. Both sequences are bounded below. If we assume that both sequences are also decreasing, we can apply Monotone Convergence Theorem to state that both are convergent to some limit $l$ (all subsequences has the same limit as sequences they are subsequences of). Now, since $b^{1/n^2}$ is subsequence of $b^{1/n}$ and $b^{1/n^2} = b^{1/n} \cdot b^{1/n}$, we can apply theorem about product of sequence limits to get $l = l \cdot l \ge 1$, which gives us only $1$ as possible limit value. The proof for $1 \gt b \gt 0$ is similar, but inequalities are reversed and we assume sequence is increasing.
The main problem I have with this proof is "jump" from $b^{1/n} \gt 1$ to $\lim_{n \to \infty} b^{1/n} \ge 1$. Feedback on improving it, if possible, is welcome.
"Now, since $b^{1/n^2}$ is subsequence of $b^{1/n}$ and $b^{1/n^2} = b^{1/n} \cdot b^{1/n}$, we can apply theorem about product of sequence limits to get $l = l \cdot l \ge 1$, which gives us only $1$ as possible limit value."
This statement has a few flaws, as you stated yourself, how can you arrive to the fact that $l\cdot l\ge 1$ and even if that is true it doesn't necessarily mean that $l=l\cdot l$, nor does it mean that $l=1$. For example since $l>1$, we can also say $l\ge 0$ yet that obviously doesn't imply $l=0$.
I would do this a bit differently, while still using Monotone Convergence Theorem. Based on your assumption that $b^{\frac{1}{n}}$ is (monotonically ) decreasing (which, keep in mind you should usually prove) we only need to show that this sequence for $b>1$ is bounded below by $1$. Let's take some arbitrary $\varepsilon>0$, by definition of infimum of a sequence we have that $\displaystyle b^{\frac{1}{n}}<1+\varepsilon\implies n>\log_{1+\varepsilon}b$.
Which means $\forall\varepsilon>0 \exists n>\log_{1+\varepsilon}b$, such that $\displaystyle b^{\frac{1}{n}}<1+\varepsilon$.
Similarly you can show for $0<b<1$, but the sequence is, instead, increasing and you should be finding the supremum of it instead.