I'm trying to solve this problem:
Suppose that $\lim_{x \rightarrow c} f(x)$ exists. Prove that there exists a constant $M$ and a $\delta > 0$ such that $|f(x)|<M$ for $0 < |x - c| < \delta $.
My attempt is as follows:
From the $(\varepsilon, \delta ) $-definition of the limit, we have that for every $\varepsilon > 0$, there exists a $\delta$ such if $0 < |x - c | < \delta$, then $|f(x) - L| < \varepsilon$. Then from the definition of the absolute value we have that $-\varepsilon < f(x) - L < \varepsilon$, and so $L-\varepsilon < f(x) < L+\varepsilon$. Furthermore since $-L - \varepsilon < L- \varepsilon$, we have that $-L - \varepsilon <f(x) < L+ \varepsilon$ or $|f(x)| < L + \varepsilon = M$ as required.
I'm specifically wondering if the part where I said $-L - \varepsilon < L- \varepsilon$ is valid, since $L$ could be negative?
As you say, you cannot do that step like that since $L$ could be negative. What you can do is, from $L-\varepsilon<f(x)<L+\varepsilon$ you get $$ |f(x)|\leq\max\{|L-\varepsilon|,|L+\varepsilon|\}\leq |L|+\varepsilon. $$
Small criticism: you don't need an "arbitrary" $\varepsilon$. Just take $\varepsilon=1$ (say) to get $|f(x)|\leq|L|+1$.
A more natural way to do the problem: use the (reverse) triangle inequality. So you have $$ |f(x)|-|L|\leq|f(x)-L|<\varepsilon, $$ and thus $|f(x)|\leq |L|+\varepsilon$.