If $Y=\min\{ X_{1},X_{2} \}$ and $X_{i} \sim N(0,1)$ then $Y^{2} \sim \chi ^{2}_{(1)}$ If $Y=\min\{ X_{1},X_{2} \}$ and $X_{i} \sim N(0,1)$ then $Y^{2} \sim \chi ^{2}_{(1)}$
If $X \sim  N(0,1) \implies X^{2} \sim \chi ^{2}_{1}$
What about the square?
If we suppose that $X_{1} < X_{2}$ we have the previous proposition and is the same for $X_{2} < X_{1}$.
 A: If $X_1,X_2 \stackrel{i.i.d}{\sim} N(0,1)$ and $Y = \min(X_1,X_2)$, then
$Pr(Y > y) = \prod^{2}_{i=1} Pr(X_i > y)=(1-\phi (y))^2 \implies F_Y(y) = 1 - (1-\phi (y))^2$, so
$f_Y(y) = 2(1-\phi(y))\phi'(y)$
Let $T = Y^2$, then
$Pr(T\leq t) = Pr(Y^2 \leq t) = Pr(-\sqrt{t} \leq Y \leq \sqrt{t}) = F_Y(\sqrt{t})-F_Y(-\sqrt{t})$
$ \implies f_T(t) = \frac{1}{\sqrt{t}} (f_Y(\sqrt{t}) - f_Y(-\sqrt{t})$, so
$f_T(t) = \frac{\phi'(\sqrt{t})}{\sqrt{t}}$ for $t>0$, which is the pdf of chi-square with 1 df.
A: The intuition is correct, but you could provide an analytical solution without having to differentiate at all. Start by obtaining the distribution of $Y$:
$$
1-F_{Y}(y) = P\left(\min\left\{X_{1},X_{2}\right\}\geq y\right) = P\left(X_{1}\geq y,X_{2}\geq y\right) = P\left(X_{1}\geq y\right)P\left(X_{2}\geq y\right) = \left(1-\Phi(y)\right)^{2}
$$
where $\Phi(y)$ is the standard normal distribution. Hence
$$
F_{Y}(y) = 1- \left(1-\Phi(y)\right)^{2} = 1-\Phi(-y)^{2}
$$
Now let $Z\equiv Y^{2}$. Let's compute its distribution
$$
F_{Z}(z) = P\left(Y^{2}\leq z\right) = P\left(-\sqrt{z}\leq Y\leq \sqrt{z}\right) = F_{Y}\left(\sqrt{z}\right)-F_{Y}\left(-\sqrt{z}\right) = \Phi\left(\sqrt{z}\right)^{2}-\Phi\left(-\sqrt{z}\right)^{2}
$$
Now use the formula for the difference between squares
$$
F_{Z}(z) = \underbrace{\left(\Phi\left(\sqrt{z}\right)+\Phi\left(\sqrt{z}\right)\right)}_{=1}\left(\Phi\left(\sqrt{z}\right)-\Phi\left(-\sqrt{z}\right)\right) = P\left(-\sqrt{z}\leq X\leq \sqrt{z}\right)=P\left(X^{2}\leq z\right)
$$
for $X\sim\mathcal{N}(0,1)$, completing the proof.
