real analysis - prove that a function $f$ is continuous at $x_0$ iff $f(x_{0}^{-}) = f(x_{0}^{+}) = f(x_0)$

prove that a function $f$ is continuous at $x_0$ iff

$$\lim\limits_{x \rightarrow x_{0}^{-}} f(x) = \lim\limits_{x \rightarrow x_{0}^{+}} f(x) = \lim\limits_{x \rightarrow x_0}f(x)$$

By definition, a function $f$ is said to be continuous at $x=x_0$ if $\lim\limits_{x \rightarrow x_{0}} f(x) = f(x_0)$

It follows that for each $\epsilon>0$ there is a $\delta>0$ s.t.

$$| f(x) - f(x_0)|< \epsilon$$ whenever $$|x-x_0|<\delta$$

Considering

$$|x-x_0|<\delta$$ is true
$$-\delta < x -x_0 < \delta$$
$$x_0 - \delta < x < x_0 + \delta$$

Therefore,
On the right of $x_0$, it follows that : $x_0 \leq x < x_0 + \delta$ such that $| f(x) - f(x_0)|< \epsilon$.
It shows that the function f is continuous from the right at $x_0$,
Therefore:

$$\lim\limits_{x \rightarrow x_{0}^{+}} f(x) = \lim\limits_{x \rightarrow x_0}f(x)$$

Similarly, on the left of $x_0$, it follows that $x_0 - \delta < x \leq x_0$ such that $| f(x) - f(x_0)|< \epsilon$.
It shows that the function f is continuous from the left at $x_0$.
Therefore:

$$\lim\limits_{x \rightarrow x_{0}^{-}} f(x) = \lim\limits_{x \rightarrow x_0}f(x)$$

We can conclude then that if $f$ is continuous at $x_0$, $f$ has to be continuous from both the left and the right at $x_0$,It follows that

$$\lim\limits_{x \rightarrow x_{0}^{-}} f(x) = \lim\limits_{x \rightarrow x_{0}^{+}} f(x) = \lim\limits_{x \rightarrow x_0}f(x)$$

Is this correct? anything could have been done differently? Any input is much appreciated