Continuity and limits. Please check epsilon delta 
Suppose $f$ is continuous at $a$ and $f(a) = 0$. Prove that if $\alpha \neq 0 $, then $f+\alpha$ is nonzero in some open interval containing $a$.

Since $f$ is continuous, we take $\epsilon = |\alpha|$; then we have $|x - a| < \delta \implies -|\alpha| < f(x) < |\alpha| \implies \alpha - |\alpha | <f(x) + \alpha < \alpha + |\alpha |$
We consider two cases. If $\alpha >0$, then $\alpha = |\alpha|$; thus  $|x - a| < \delta \implies 0<f(x) + \alpha < 2|\alpha |$
On the other hand, if $\alpha < 0$ we have $|x - a| < \delta \implies -2|\alpha |< f(x) + \alpha < 0$
EDIT Spivak's answer book referred me back to a theorem in the book. I am unfortunately too lazy to do that and after a quick glimpse of the theorem, it didn't look anything like what I wrote...
 A: Correct. Note that in such situations, people usually give themselves some space and take $\epsilon=|\alpha|/2$. And it is not recommended to take the habit to rely too much on strict inequalities. It does no matter in this case, of course. But in more involved arguments requiring a limiting process, strict inequalities become large...
Also, you can handle both cases at once with the reverse triangular inequality:
$$|t-z|=|z-t|\geq ||z|-|t||\geq |z|-|t|.$$ 
So take $\delta>0$ such that $|x-a|\leq \delta$ implies $|f(x)|\leq \frac{|\alpha|}{2}$. Then
$$
|f(x)+\alpha|\geq |\alpha|-|f(x)|\geq |\alpha|-\frac{|\alpha|}{2}=\frac{|\alpha|}{2}>0.
$$
Note: the theorem you mention could very well be that if $f$ and $g$ are continuous at $a$, then $f+g$ is continuous at $a$. In this case, take $g$ the constant function equal to $\alpha$, which is obviously continuous. So $h=f+\alpha$ is continuous at $a$ where it takes the value $\alpha\neq 0$. Hence $|h|$ is continuous at $a$ where it takes the value $|\alpha|>0$. Take $\epsilon=\frac{|\alpha|}{2}>0$. Then there is $\delta>0$ such that $|x-a|\leq \delta$ implies
$$
0<\frac{|\alpha|}{2}=|\alpha|-\frac{|\alpha|}{2}\leq |h(x)|=|f(x)+\alpha|\leq |\alpha|+\frac{|\alpha|}{2}
$$
where the rhs is useless for your purposes.
