Prove/Disprove:
Let $A\subseteq B\subseteq \mathbb{R}$ If $B$ is bounded from below so $A$ is bounded from below.
$B$ is bounded from below $\rightarrow$ $x\leq B$.
Let there be $z\in A$, we know that $A\subseteq B$ so $z\in B$ therefore $x\leq z: \forall z\in B$ So $x\leq z$ and $A$ is bounded from below too.
Is the logic correct?
On
Your logic is mostly correct; there are a few details that I would include in your proof that may seem trivial, but with mathematical proofs, oftentimes small details are the difference between a right and a wrong answer.
I would say:
Let there be a value $z\in A$. By hypothesis, since $A\subset B$, we know that $z\in B$ as well. If $B$ is bounded below, then there exists a value $b$ for which $b\leq x$ for every $x\in B$. Thus, we know that $b\leq z$ for all $z\in A$ as well (in other words, since all of the $z$ values are a subset of the $x$ values, and $b\leq x$ for all $x$, it is only logical that $b\leq z$ for those $x$ values that also happen to be $z$ values). The bolded statement is the definition of a lower bound, so $A$ is also bounded from below. Q.E.D.
Hope that makes sense!
For the sake of not leaving the question unanswered:
It looks good but little correction at the end:
.. so $z\in B$ therefore $x\leq z, \forall z\in A$ and hence $A$ is bounded from below.