Can you help me check if this proof is correct? If not, kindly provide a better proof
Prove that if $f$ is continuous at $x_0\in [a,b]$ and $f(x_0)\neq 0,$ then $\sup\limits_{P}L(|f|,P)>0$
Suppose $a<b$. For $\epsilon=|f(x_0)|/2,$ there exists $\delta>0$ such that $$|f(x)|>\dfrac{1}{2}|f(x_0)|,\;\;\text{whenever}\;\;x\in(x_0-\delta,x_0+\delta).$$ Choose a uniform partition $P_n$, for each $n,$ such that $$a=x_0<x_1<\cdots<x_n=b\;\;\text{and}\;\;x_j-x_{j-1}=(b-a)/n,\;\;j\in \{1,2,\cdots,n\}.$$ Hence, \begin{align}\sup\limits_{P_n}L(|f|,P_n)&= \lim\limits_{n\to \infty}\sum^{n}_{j=1}|f(t_j)|(x_j-x_{j-1})\\&\geq \dfrac{1}{2}|f(x_0)|\lim\limits_{n\to \infty}\sum^{n}_{j=1}(x_j-x_{j-1})\\&\geq \dfrac{1}{2}|f(x_0)|\lim\limits_{n\to \infty}(x_n-x_{0})\\&= \dfrac{1}{2}|f(x_0)|\lim\limits_{n\to \infty}(b-a)\\&= \dfrac{1}{2}|f(x_0)|(b-a)\\&>0\end{align}
No, it is not correct. That inequality containg the second $\geqslant$ doesn't hold; you seem to be assuming here that each $m_j$ is greater than or equal to $\frac12\bigl\lvert f(x_0)\bigr\rvert$, but that is not true.
Note that you only have to proved an example of one partition $P_0$ such that $L\bigl(\lvert f\rvert,P_0\bigr)>0$. Then it will follow automatically that $\displaystyle\sup_PL\bigl(\lvert f\rvert,P\bigr)>0$.