Suppose $(a,b)$ is an open and bounded interval of $\mathbb{R}$. Also, let $$f:(a,b)\to \mathbb{R}$$ be a bounded function. Set $\tilde{f}:[a,b]\to\mathbb{R}$ as the function $$\tilde{f}(x)=\begin{cases} f(x)\qquad x \in (a,b)\\\alpha \qquad x=a\\\beta \qquad x=b\end{cases}.$$ Is it true that if $\tilde{f}$ is Riemann integrable over $[a,b]$ then the improper integral of $f$ over $(a,b)$ converge and $$\int_{a}^{b}f=\int_{a}^{b}\tilde{f}\qquad ?$$
Note that the improper integral of $f$ over $(a,b)$, provided that $f$ is riemann integrable on any compact subinterval $[c,d]\subset (a,b)$, is defined as $$\int_{a}^{b}f=\lim_{m\to a^+}\int_{m}^{c}f+\lim_{M\to b^-}\int_{c}^{M}f$$ where $c\in (a,b)$ is arbitrary.
Personal attempt Since $\tilde{f}$ is riemann integrable over $[a,b]$ then it's riemann integrable over any $[c,d]\subset (a,b)$. But $\tilde{f}=f$ on such compact intervals. So $f$ is riemann integrable over any $[c,d]\subset (a,b)$. Then, for any fixed $c\in (a,b)$ one has $$\int_{a}^{b}f(t)dt=\lim_{m\to a^+}\int_{m}^{c}f(t)dt+\lim_{M\to b^-}\int_{c}^{M}f(t)dt=\lim_{m\to a^+}\int_{m}^{c}\tilde{f}(t)dt+\lim_{M\to b^-}\int_{c}^{M}\tilde{f}(t)dt=\int_{a}^{c}\tilde{f}+\int_{c}^{b}\tilde{f}=\int_{a}^{b}\tilde{f}.$$
Your proof is correct except for the fact that you need to justify the last step (the use of second last '$=$' sign). It is based on the Fundamental Theorem of Calculus a part of which says that if a function $f:[a, b] \to\mathbb {R} $ is Riemann integrable on $[a, b] $ and if $F(x) =\int_{a} ^{x} f(t) \, dt$ then the function $F$ is continuous on $[a, b] $.
This also shows that the notion of improper integral for bounded functions on bounded intervals is superfluous (but perhaps attractive to many as some questions on this website suggest). The usual (or shall we say proper) Riemann integral is powerful enough to deal with bounded functions on bounded intervals and this is a precondition for the definition of Riemann integral. For example the integral $\int_{0}^{1}\sin(1/x)\,dx$ is not improper but $\int_{0}^{1}(1-x^2)^{-1/2}\,dx$ is improper (and important at the same time).