If $f$ is uniformly continuous on $[a,b]$ it is uniformly continuous on $(a,b)$ as well. If $f$ is uniformly continuous on $(a,b)$ then that means $f$ has a continuous extension to the interval $[a,b]$. Since this extension is continuous in a closed and bounded interval, it must be uniformly continuous in that interval.
Is this correct?
If you need a proof that "uniformly continuous $(a,b)$" implies "continuous extension on $[a,b]$", then here's one.
Let $\{x_n = a+1/n\}\subset (a,b)$. Let $\varepsilon>0$ arbitrary. Then there exists $\delta>0$ such that if $|x-y|<\delta$, then $|f(x)-f(y)|<\varepsilon$. Let $N>2/\delta$, if $m,n>N$, then $$ |x_n-x_m| = |a-1/n+b-1/m| = |1/n-1/m|\leq 1/n+1/m \leq \delta. $$ Note that this holds for all natural numbers $m,n>N$. Therefore by absolute continuity $$ |f(x_n)-f(y_n)|<\varepsilon. $$ This meas that $\{f(x_n)\}$ is a Cauchy sequence, that it, it admits a limit. Let's call $f(a) = \lim_n f(x_n)$. We want to show that $f$ is continuous there. So let $\varepsilon>0$ arbitrary. Now choose $\delta>0$ so that $|f(x)-f(y)|<\varepsilon$ for $|x-y|<\delta, x,y\in (a,b)$. Pick any $z,x_n\in (a,a+\delta)$, $z$ being arbitrary and $x_n$ from the sequence above. We can estimate $$ |f(a)-f(z)|\leq |f(a)-f(x_n)|+ |f(x_n)-f(z)| \leq |f(a)-f(x_n)|+\varepsilon. $$ Pass to the limit in the above inequality to obtain $$ \lim_n |f(a)-f(z)| = |f(a)-f(z)|\leq \lim_n |f(a)-f(x_n)|+ \varepsilon = \varepsilon.$$ That is, the extension for $f$ is continuous at $a$. Analogous for $b$.
At this point the extension is continuous on a bounded, closed interval, i.e. is uniformly continuous (classical result). This established the equivalence of uniform continuity over $(a,b)$ or $[a,b]$.