I can perfectly understand the $(\Rightarrow)$ part, but the the rest seems to be a bit harder. I get the idea, but I'm still not convinced I understood it. This is my work from the beginning:
First part:
Suppose $\lim_{x\to a} = L$, so given $\varepsilon \gt 0$ there exists $\delta \gt 0$ such that $0 \lt |x - a| \lt \delta \implies |f(x) - L| \lt \varepsilon$.
If $x \in (a-\delta, a)$, so $x \in (a - \delta, a + \delta)$, then $\lim_{x\to a^-} = L$
if $x \in (a, a + \delta)$, so $x \in (a - \delta, a + \delta)$, then $\lim_{x\to a^+} = L$
Thus $\lim_{x\to a} \implies \lim_{x\to a^-} = L = \lim_{x\to a^+}$.
Second part:
Suppose $\lim_{x\to a^-} = L = \lim_{x\to a^+}$. Let $\varepsilon$ be given.
Considering $\lim_{x\to a^-} = L$, so there exists $\delta_1 \gt 0$ such that $a - \delta_1 \lt x \lt a \implies |f(x) - L| \lt \varepsilon$. Likewise $\lim_{x\to a^+} = L$ so there exists $\delta_2 \gt 0$ such that $a \lt x \lt a + \delta_2 \implies |f(x) - L| \lt \varepsilon$. This is now the part that gets me.
Set $\delta = $ min{$\delta_1 , \delta_2$}. Let's say $\delta_1$ is the smallest, so $0 \lt |x - a| \lt \delta_1 \implies a - \delta_1 < x < a + \delta_1$. Now $x \in (a - \delta_1, a + \delta_1)$, does it automatically implies that $x \in (a - \delta_2, a + \delta_2)$? So now $\lim_{x\to a^-} = L = \lim_{x\to a^+}$ $\implies$ $\lim_{x\to a} = L$?
I read a proof before, but this last part wasn't clear, the way it was done was a bit confusing to me.