The correspondence theorem to which I refer is the bijection between ideals of a commutative ring with $1$, $A$, and ideals of $A/\frak{b}$. I can prove easily most parts that imply the bijection except for this last bit: $\pi^{-1} \pi (\frak{a}) \subset \frak{a}$, where $\pi$ is the natural surjection $A \to A/\frak{a}$.
Attempt: suppose that $\pi^{-1} \pi(\frak{a} ) \supsetneq \frak a$. Take $x \in$ the difference. Then $x + \frak{a}$ is a coset disjoint from & not equal to $0 + \frak a$ in the quotient ring. Also notice that $x = x + 0 \in x + \frak{a} \subset \pi^{-1}\pi(\frak{a}) $. Composing the first set inequality with $\pi$ then gives us $\pi(\frak{a}) \supsetneq \pi(\frak{a})$ since $x + \frak{a} \not\subset \frak{a}$. Contradiction.
Is this proof correct? How can I make this proof better or more elegant?
Here's how I'd rephrase your argument (and prove the bijection directly all at once):
$$\begin{align*} \pi^{-1}(\pi(\mathfrak{a}))&=\pi^{-1}(\{\pi(x):x\in\mathfrak{a}\})\\[0.05in] &=\pi^{-1}(\{x+\mathfrak{a}:x\in\mathfrak{a}\})\\[0.05in] &=\pi^{-1}(\{0+\mathfrak{a}\})\\[0.05in] &=\{y\in A:\pi(y)=0+\mathfrak{a}\}\\[0.05in] &=\{y\in A: y+\mathfrak{a} = 0+\mathfrak{a}\}\\[0.05in] &=\{y\in A: y\in\mathfrak{a}\}\\[0.05in] &=\mathfrak{a} \end{align*}$$