Here is Prob. 12, Sec. 5.2, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
A function $f \colon \mathbb{R} \to \mathbb{R}$ is said to be additive if $f(x+y)= f(x) + f(y)$ for all $x, y$ in $\mathbb{R}$. Prove that if $f$ is continuous at some point $x_0$, then it is continuous at every point of $\mathbb{R}$.
My Attempt:
Let $c$ be an arbitrary real number.
First suppose that $x_0 = 0$. As $f$ is continuous at the point $0$, so given any real number $\varepsilon > 0$ we can find a real number $\delta > 0$ such that
$$\lvert f(x) \rvert = \lvert f(x) - 0 \rvert = \lvert f(x) - f(0) \rvert < \varepsilon $$
for all $x \in \mathbb{R}$ for which
$$ \lvert x \rvert = \lvert x-0 \rvert < \delta. $$
Then
$$ \lvert f(x) - f(c) \rvert = \lvert f( x - c ) \rvert < \varepsilon $$
for all $x \in \mathbb{R}$ such that
$$ \lvert x-c \rvert < \delta. $$
Thus it follows that $f$ is continuous at every point $c \in \mathbb{R}$.
Am I right?
Now let us suppose that $x_0 \neq 0$. As $f$ is continuous at $x_0$, so, for every real number $\varepsilon > 0$ we can find a real number $\delta > 0$ such that
$$ \left\lvert f \left(x - x_0 \right) \right\rvert = \left\lvert f(x) - f \left( x_0 \right) \right\rvert <
\varepsilon $$
for all real numbers $x$ which satisfy
$$ \left\lvert x- x_0 \right\rvert < \delta. $$
Now if we could show from here that $f$ is continuous at $0$, then as before we will have managed to show that $f$ is continuous at every real number $c$.
What next? How to proceed from here to show that our function $f$ is continuous at $0$? Or, is there any other route?
Answer
You are almost there. Let $| t-0 | = |t| < \delta$, where $\delta$ is the one you have in the second box. Then $t + x_0 = x\in \mathbb{R}$, so $|t| = |x - x_0| < \delta$.
Thus, $|f(t) - f(0)| = |f(t)| = |f(x-x_0)| = |f(x) - f(x_0)| < \epsilon$ for $|t| < \delta$.
So $f$ is continuous at $0$.
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