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I asked a question similar to this previously but I realized that what I was trying to prove was false. Then I changed the thing I was trying to prove but it was still false. I also accepted an answer that proved my theorem in a special case but that was not true in general. I am trying to prove it again and people stopped commenting on my previous question because I accepted the answer.

So I am asking a new question.

The correct version of the theorem I want to prove is "Suppose that $f$ is continuous on $[a,b]$, $f(a)<0$, and $f(b)>0$. Then $exists cin (a,b), exists delta>0, forall xin (c-delta ,c), f(x)le 0$ and $forall xin (c,c+delta ), f(x)ge 0$.

I tried to prove this by contradiction. I assumed that $forall cin(a,b), forall delta>0, exists xin(c-delta,c), f(x)>0$ or $exists xin(c-delta,c), f(x)<0$.

By Strong IVT I know that $exists a_1in (a,b), forall xin (a,a_1), f(x)<0$.
Also by Strong IVT I know that $exists b_1in (a,b), forall xin (b_1,b), f(x)>0$.
I see that $f(a_1)=0$ and $f(b_1)=0$.

I also know to prove that $a_1$ satisfies what I want if $a_1$ is an isolated zero.

Similarly, if $b_1$, I know to finish the proof (in both cases I don't need a proof by contradiction).

Then there is the case where neither $a_1$ or $b_1$ are isolated. Everything I write past this is speculation. Because $f$ is continuous on $[a,b]$ and $[a,b]$ is compact, I know that $f$ is uniformly continuous on $[a,b]$. Which means that I know that $varepsilon>0$, $exists delta>0$, $forall x,yin[a,b]$, $|x-y|<delta Longrightarrow |f(x)-f(y)|<varepsilon$.

What I tried to do is "walk" from $a_1$ to $b_1$ to show that for some point $x$ in $(b_1,b) f(x)<0$. I attempted this with the following (insufficient) process.
I chose $delta_n$ such that $forall xin[a,b], |x-y|<delta_n Longrightarrow |f(x)-f(y)|<-f(a)cdot2^{-n}$. Then I defined a sequence $x_n$ such that $x_1=a_1$ and $x_n=x_{n-1}+text{min}left{frac{1}{2}delta_n,(b-a_1)2^{-(n+1)}right}$. The first term in the minimum ensures that $f(x_n)ge -f(a)cdot2^{-n}>0$ while the second term prevents $x_n$ from leaving the interval $[a,b]$.
The problem I see with this attempt is that if $x_n$ does not enter $(a,b)$ for a finite $n$ (which would occur if $delta_n$ were too small) then I wouldn't get a contradiction.

Perhaps the correct approach is to use the fact that I am considering the case where $a_1$ and $b_1$ are not isolated. If someone can show me how to prove that $f(x) = |x| sin(1/x)$ if $xne 0$ and $f(0)=0$ satisfies "$exists cin (-1,1), exists delta>0, forall xin (c-delta ,c), f(x)le 0$ and $forall xin (c,c+delta ), f(x)ge 0$ or give me a hint for such a proof then I think I can apply similar reasoning to the general case.

I know that it is possible to show that this holds by finding a $c$ where $f(c)=0$ and $f'(c)>0$, but I want to prove this without assuming that my function is differentiable.

Unfortunately I believe your proposition is false. Your question is intimately connected to this question:

Continuous function changing sign on Cantor set

The basic issue is that there is a continuous function whose set of points at which it changes sign is contains no isolated points which would be required in order for your proposition to be true.