Assume $f(t) \ge 0$ for all $t$ with $f(0)>0$ and as long as $f(t)>0\,(f(t) \neq 0)$, it is continuous at that $t$. Also assume there is a $T>0$ such that $f(T)=0$.
Does $\min \{t\in[0,T]\,|\,f(t)=0\} $ exist?
The set is non-empty because $f(T)=0$. So if I show the above set is closed (and hence compact because it's bounded) then the minimum exists. But how can we show the set is closed? Again If I knew $f$ is always continuous then $f^{-1}(0)$ is closed and everything is done but I don't have the assumption that $f$ is continuous on $[0,T]$. The only assumption I have is that $f\ge 0$ and as long as $f(t)>0$ it is continuous at that $t$. Is the information enough to show the above minimum value exists or there is a counter-example for this question?
I faced this particular question in a much bigger proof that I am doing for a particular problem. So I appreciate any help!
The conditions as stated are not sufficient.
As a counterexample, let $T > 1$, and let $f$ be given by $$ f(t)= \begin{cases} 1&&\text{if}\;\,t < 1\\[4pt] -1&&\text{if}\;\,t=1\\[4pt] 0&&\text{if}\;\,t > 1\\ \end{cases} $$ Update:
With your latest edit, the conditions are sufficient.
By the continuity condition, if $t$ is such that $f(t) > 0$, then $f$ is positive in some open interval containing $t$.
It follows that $$\{t{\,\mid\,}f(t) > 0\}$$ is an open set.
Then assuming $f$ is everywhere nonnegative, it follows that $$\{t{\,\mid\,}f(t) = 0\}$$ is a closed set.
Then since $f(T)=0$ for some $T > 0$, it follows that $$\{t\in[0,T]{\,\mid\,}f(t) = 0\}$$ is compact and nonempty, hence has a least element.