If $f$ is uniformly continuous on $(a, b]$ and $[b, c)$ prove that it is uniformly continuous on $(a, c)$.
Assuming the contrary f isn't uniformly continuous in $(a,c)$ then there exists a sequence $x_k, y_k$ and an $\epsilon_0$ such that $$|x_n-y_n|\rightarrow0 \quad \text{and} \quad|f(x_n)-f(y_n)|\ge\epsilon_0$$ Consider the subsequence $x_{n_k}, y_{n_k} \in(a,b]$ then $|f(x_{n_k})-f(y_{n_k})|\rightarrow 0$ as it is uniformly continuous in this interval. This contradicts $|f(x_n)-f(y_n)|\ge\epsilon_0$.
Hence it is uniformly continuous.
Is this proof ok?
I think that your proof is not complete. What happens when the sequence $(x_n)_n$ is contained in $(a,b]$ and the sequence $(y_n)_n$ is contained in $[b,c)$?
Then $x_n\to b^-$ and $y_n\to b^+$ and, by the continuity of $f$ at $b$, it follows that $|f(x_n)−f(y_n)|\to 0$.
P.S. Note that this is also the reason why the uniform continuity of $f$ on $(a, b]$ and on $(b, c)$ does not imply the uniform continuity on $(a, c)$