The forward direction is simple. Assume that $\lim_{x \to x_0}f(x) = L$ . Then given $\epsilon$ there is a $\delta$ such that \begin{equation*} |f(u)- L| < \epsilon/2 \text{ when } 0 < |u-x_0| < \delta. \end{equation*} Now for $0 < |u - x_0 | < \delta$ and $ 0< |v- x_0 | < \delta$ we have \begin{equation*} |f(u)- L| < \epsilon/2 \text{ and } |f(v)-L| < \epsilon/2 \end{equation*}and so

\begin{equation*} |f(u)-f(v)| = | f(u)- L -(f(v)-L)| \leq |f(u)-L|+|f(v)-L| < \epsilon/2+\epsilon/2 = \epsilon. \end{equation*} We prove the reverse by contradiction. Assume that the condition holds. Now suppose that $\lim_{x \to x_0}f(x)$ does not exist. This means that for any $l$ and any $\epsilon$ sufficiently small then for any $\delta>0$ there is $x_l$ such that $0<|xl-x_0|< \delta~{and}~|f(x_l)-l| \geq \epsilon$ . For any such $\epsilon$ choose $u$ such that $0 < |u-x_0| < \delta $ and put $l=f(v)$ then substituting in the condition with $u=x_l$ we get $|f(x_l)-l| < \epsilon$ . A contradiction.