$\varepsilon-\delta$ argument to show that uniform continuity of a real-valued function $f$ on some interval $A$ implies that $f$ is continuous on $A$ Here is an $\varepsilon-\delta$ argument to show that uniform continuity of a real-valued function $f$ on some interval $A$ implies that $f$ is continuous on $A$.

Suppose $f$ is a real-valued function that is uniformly continuous on the interval $A$. This means that for any $\varepsilon$, there is a corresponding $\delta_{\varepsilon}$ such that for any $x,y \in A$ we have that if $|x-y|\lt \delta_\varepsilon$, then $|f(x)-f(y)|\lt \varepsilon \quad (\dagger_1)$.
We will now show that at any arbitrary $a \in A$, $f$ is continuous.
Consider some arbitrary $a$ and some arbitrary $\varepsilon$. Consider the $\delta_{\varepsilon}$ that $(\dagger_1)$ supplies us with. Next, consider the open interval $\left(a-\frac{\delta_{\varepsilon}}{2},a+\frac{\delta_{\varepsilon}}{2}\right)$. Choose any $x,y$ in this interval: then we have that $a-\frac{\delta_{\varepsilon}}{2} \lt x \lt a+\frac{\delta_{\varepsilon}}{2}$ and $a-\frac{\delta_{\varepsilon}}{2} \lt y \lt a+\frac{\delta_{\varepsilon}}{2}$. Some manipulation of the inequalities will show that $x-y \lt \delta_{\varepsilon}$ and $-\delta_{\varepsilon} \lt x-y$, which implies that $|x-y| \lt \delta_{\varepsilon}$.
Then an application of $(\dagger_1)$ tells us that for any $x,y \in  \left(a-\frac{\delta_{\varepsilon}}{2},a+\frac{\delta_{\varepsilon}}{2}\right)$, we must have $|f(x)-f(y)|\lt \varepsilon$. Consider the particular instance of $y:=a$. Then we have that for any $x \in \left(a-\frac{\delta_{\varepsilon}}{2},a+\frac{\delta_{\varepsilon}}{2}\right): |f(x)-f(a)| \lt \varepsilon$.
But this is the definition of continuity...in particular, the desired $\delta$ for this $\varepsilon$ is $\frac{\delta_{\varepsilon}}{2}$.

Edit: I think I need to be more careful with the interval $\left(a-\frac{\delta_{\varepsilon}}{2},a+\frac{\delta_{\varepsilon}}{2}\right)$ to ensure that this sits inside $A$.

Although I see no errors, it is clear to me that it's much simpler than you think it is. Given $\varepsilon>0$, if $|x-a|<\delta_\varepsilon$, and if $x\in A$, then $\bigl|f(x)-f(a)\bigr|<\varepsilon$, and therefore $f$ is continuous at $a$. That's all.
