real analysis - Provided $f$ is continuous at $x_0$, and $f(x+y) = f(x) + f(y)$ prove $f$ is continuous everywhere.

My attempt...

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By definition, whenever $|x- x_0| < \delta$ we have $|f(x) - f(x_0)| < \epsilon$. Observing that

\begin{align}
|f(x) - f(y)| &= |f(x -x_0 + x_0) + f(y)| = |f(x-x_0) + f(x_0) - f(y)| \newline \newline
&\leq \epsilon + |f(y) - f(x_0)| = \epsilon + |f(y-x_0)|...
\end{align}

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Here I need to choose a delta that can depends on $\epsilon$ and $y$ s.t. whenever $0<|x-y|< \delta$ then the above inequality is bounded by any $\epsilon$.

I'm also, in general, having trouble understanding this concept of continuity on an interval. I believe the structure of the definition is: for any $\epsilon> 0$ and any number $y$ in the interval, there exists a $\delta$ that depends on $\epsilon$ and $y$ such that for all $x$ in the interval and $|x - y | < \delta$ then $|f(x) - f(y)| < \epsilon$.

This definition makes me tempted to just choose y to be in the same delta neighborhood as $x$ in the given statement, but that constricts continuity to a small interval.

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Edit: This question assumes no knowledge of Lebesgue measure theory.