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I am trying to solve the following exercise in Rudin's "Rudin's Principles of Mathematical Analysis" book: (Ex 4.1)

Suppose $f$ is a real function defined on $R^1$ which satisfies
$$lim_hto 0[f(x+h)-f(x-h)]=0$$
for every $xin R^1$. Does this imply that $f$ is continuous?

The answer to this question is simply no, and it can be proved by using the function $f(x) = 1$ if $xin mathbbZ$ and $f(x) =0$, otherwise.

However, I am a little bit confused by this result, since I obtain the contrary by using the definitions of limits and continuity. In particular, I have the following derivation:

Define the function $$g(h) := f(x+h) - f(x-h)$$ for a fixed x. Then the hypothesis implies that $$lim_hto 0g(h)=0.$$ By using the definition of the limit, we have that $forall varepsilon >0$, $exists delta>0$, such that $|g(h)| < varepsilon$ and $|h|< delta/2$.

Now let $p, q in mathbbR$. Assume $p < q$, without loss of generality. Define $h := fracq-p2$, $x := fracq+p2$. Fix $varepsilon >0$. Then, for $$|f(q)- f(p)| = |f(x+h)-f(x-h)| < varepsilon,$$
we know that
$$|q-p| = |x+h-x+h| = 2 |h| < delta.$$ Hence, $f$ is continuous.

My question is: in which part of this derivation I am making a mistake?

Please note that there have been other questions regarding this exercise, for instance see this. However, I have a different concern, whose solution I think could be useful for other people.

The error lies in “Hence, $f$ is continuous.” There is nothing before that assertion that justifies it. Proving that $f$ is continuous means proving that, for every $xinmathbb R$ and every $varepsilon>0$, there is a $delta>0$ such that $lvert hrvert<deltaimpliesbigllvert f(x+h)-f(x)bigrrvert<varepsilon$, and you did not prove that.

I think what is confusing you is that you showed something that looks like continuity, but is in fact not. To show that $f$ is continuous, we would have to show that $f$ is continuous at $x$ for each $x$. That would involve showing that $pmbf(x+h)-f(x)$ can be made arbitrarily small by taking $h$ sufficiently small.

This is what your proof does not show: instead you demonstrated that $pmb$ can be made arbitrarily small by taking $h$ sufficiently small, which is actually just the condition $$lim_hto0[f(x+h)-f(x-h)] = 0.$$