Set A

Let's find Solution set of $x^2 - 4 \geq 0$

$x^2 - 4\geq 0$ $(x+2)(x-2) \geq 0$ $x \in (-\infty, -2] \cup [2, \infty)$

Therefore, set $A$ is $\boxed{A = (-\infty, -2] \cup [2, \infty)}$

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Set B

Let's find Solution set of $4y^2 - 1 \geq 0$

$4y^2 - 1\leq 0$ $(2y+1)(2y-1) \leq 0$ $y \in \bigg[-\frac12, \frac12\bigg]$

Therefore, set $B$ is $\displaystyle\boxed{B = \bigg[-\frac12, \frac12\bigg] - \{0\}}$

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Proving $|A| = |B|$

Two sets are said to be equivalent if there exists a bijection $f:A \rightarrow B$ .

Now, for all $x$ in $A$ there exists $y$ in $B$ such that $\displaystyle y = \frac1x$

Also, for all $y$ in $B$ there exists $x$ in $A$ such that $\displaystyle x = \frac1y$

Therefore there exists a bijection $f:A \rightarrow B$ which is $\displaystyle f(x) = \frac1x$

Therefore $\color{#3D99F6}{\boxed{|A| = |B|}}$