Firstly, I do realize that this proof was asked about before here, but the proposed proof was different and the answers were also not complete.
To get to the point: I have written my own proof of this simple theorem, however I'm not sure if the proof is correct--the exercise is taken from Spivak, who seems to have provided different and more elaborate proof of that theorem.
My proof is following:
By the definition of limit (omitting the 'for all's and 'there exists''):
$$ \lim_{x\to a} f(x) = L_1 \implies 0 < |x - a| < δ_1 \implies |f(x) - L_1| < ε $$ and $$ \lim_{h\to 0} f(a + h) = L_2 \implies 0 < |h - 0| = |h| < δ_2 \implies |f(a + h) - L_2| < ε $$
Let's work on the second limit: $$|h| = |(a+h) - a|$$ $$\text{Let } y = a + h$$ Now, the second limit takes the following definition: $$ 0 < |y - a| < δ_2 \implies |f(y) - L_2| < ε $$ We see that both limits have the exact same form right now. By Theorem 1 [stating that if limit L exists it is necessarily unique] $L_1 = L_2$ and $\lim_{x\to a} f(x) = \lim_{h\to 0} f(a + h)$ $$Q.E.D.$$
Is my reasoning correct or have I made an error somewhere?