Why does the solution set for this inequality turn out this way? $$x^2-6x-7\ge 0$$
Steps I took:
$$(x-7)(x+1)\ge 0$$
$$x-7\ge 0\quad x+1\ge0$$
$$x\ge 7\quad x\ge -1$$
When I test the solution set, I realize that it must be $x\ge 7\quad or\quad x\le -1$
What is going on here?
 A: Up until this point 
$$(x-7)(x+1)\ge 0$$
your reasoning to correct.
But now look at the expression. It is a product of two numbers. Ask yourself: When is a product of two real numbers positive? The product is only positive, if both terms have same sign. Hence,
A)Both expressions have positive sign: $x-7\geq 0$ and $x+1\geq0$ 
or B)Both expressions have negative sign: $x-7\leq 0$ and $x+1\leq0$.
Now, look at: A) $x\geq7$ and $x\geq-1$. Both conditions are only satisfied if $x\geq7$ 
Now try to the same for B).
A: What one should do in general when solving one-variable inequalities is:


*

*Solve the equality, i.e. replace the inequality sign with an equality sign.

*Graph the function, marking the points where we have equality. 

*Now you can write down the interval where the inequality holds.

A: If $AB\ge 0$, you cannot conclude that $A\ge 0$ or $B\ge 0$. What you should be doing is something similar but different.
The function $f(x) = x^2-6x-7$ is what is called "continuous", which means that if it changes sign from $x=a$ to $x=b$, then there is a value $x=c$ between $a$ and $b$ such that $f(x)=0$. You should be looking for where the function is zero (namely at $-1$ and 7), and then testing whether $f(x)$ is positive or negative at other values.
For instance, $f(0)=-7$, so $f(x)$ is negative for all $x$ between $-1$ and 7. $f(8)=9\ge0$, so $f(x)$ is positive for all $x$ greater than 7, etc.
