Here's a series of hints as to how to attack this problem. Start with an arbitrary upper Darboux sum given by some partition $\mathcal{P}$. Each piece of the partition is an interval $[x_{i},x_{i+1}]$. What is the supremum of our function on this interval? We would like to say that it is $x_{i+1}$, since that is when the function $g(x) = x$ is biggest on the interval. Our only concern is that maybe $x_{i+1}$ is irrational, and our function is zero there. However, we know that we can find rationals in our interval arbitrarily close to $x_{i+1}$, so you should be able to prove that $x_{i+1}$ is still the supremum of $f$ on our interval, even if it is not the maximum.
Observe that, by the argument of the prior paragraph, every upper Darboux sum for $f$ is actually an upper Darboux sum for the function $g(x) = x$. What can you then say about the relationship between the upper Darboux integrals of $f$ and $g$? What do you know about the upper Darboux integral of $g$ (which is a function whose integral you can explicitly compute!)?
Here is the important theorem which may seem obvious but is difficult to prove:
Theorem: If a function $f:[a, b] \to\mathbb{R} $ is bounded on $[a, b] $ then $$\lim_{|P|\to 0}U(f,P)=\inf\,\{U(f,P)\mid P\text{ is a partition of }[a, b] \} $$ and $$\lim_{|P|\to 0}L(f,P)=\sup\,\{L(f,P)\mid P\text{ is a partition of }[a, b] \} $$
Using this theorem the job of finding the supremum of lower sums (or infimum of upper sums) is reduced to finding the limit of corresponding sums as norm of partition tends to $0$. In particular we can take a sequence of partitions which are uniform (all subintervals of equal length) such that number of subintervals tends to $\infty $.
Thus for your example it is sufficient to take the limit of $U(f, P_n) $ and $L(f, P_n) $ as $n\to\infty $. If these limits are equal (as is the case here) the function is Riemann integrable with the common limit ($b^2/2$) as its integral.
However the argument given in your question is almost correct (may be it needs a little more effort) and you shouldn't doubt yourself (not everything in analysis is difficult and even if they are you can't be wrong every time).
Here is a way to add some detail to make it perfect (from almost correct). Start with the observation that every lower sum is less than or equal to any specific upper sum. Thus for any partition $P $ we must have $$L(f, P) \leq U(f, P_n) \tag{1}$$ for all $n$ and letting $n\to\infty $ we have $$L(f, P) \leq \frac{b^2}{2}\tag{2}$$ Why does taking limit as $n\to\infty $ work for us?? Well you should further notice that as $n$ increases $U(f, P_n) $ decreases strictly and thus infimum of all $U(f, P_n) $ equals its limit $b^2/2$. And if it were possible that $L(f, P) > b^2/2$ then by definition of infimum we would have some value of $n$ such that $$b^2/2\leq U(f, P_n) < L(f, P) $$ contradicting that any lower sum can't exceed any upper sum. And thus we must have $L(f, P) \leq b^2/2$ and this justifies the process of taking limits as $n\to\infty $ and deduce equation $(2)$ from $(1)$.
In a similar manner $$U(f, P) \geq L(f, P_n) $$ gives us $$U(f, P) \geq\frac{b^2}{2}$$ Combining all the above this proves that $$L(f, P) \leq \frac{b^2}{2}\leq U(f, P') \tag{3}$$ for any partitions $P, P'$. Notice that from the result $$L(f, P_n) \leq \frac{b^2}{2}\leq U(f, P_n)$$ established in your question we can infer the stronger result $(3)$ above.
Since you have proved that the function is Riemann integrable not more than a single number can lie between all lower and all upper sums (this is also mentioned in your question which means you understand this part). What have we got now on our hands? Well, just that the integral value is $b^2/2$. Done!!
The argument presented above (in second part) can be abstracted out by stripping all details of Riemann integral and partitions to lead to a simple
Lemma: Let $A, B$ be non-empty subsets of $\mathbb{R} $ such that no member of $A$ exceeds any member of $B$. If there exist sets $C, D$ such that $$C\subseteq A, D\subseteq B, \sup C=\inf D$$ then $\sup A=\inf B$.
The sets $C, D$ can also be replaced by some sequences $x_n\in A, y_n\in B$ with $\lim x_n=\lim y_n$.
And now this looks so simple/trivial/obvious and you may easily prove it.
Best Answer
You are close to being done. Note that $x_i = a + \frac{i}{n}(b-a)$, so \begin{align*} \sum_{i=1}^{n}m_i &= \sum_{i=0}^{n-1}(a+ \frac{i}{n}(b-a)) \\ &= na + \frac{b-a}{n}\sum_{i=0}^{n-1} i \\ &= na + \frac{b-a}{n}\frac{n(n-1)}{2} \end{align*} The lower Riemann sum then becomes \begin{align*} L_f(P)=\frac{b-a}{n}\cdot\sum_{i=1}^nm_i &= a(b-a) + \frac{n-1}{n} \frac{(b-a)^2}{2} \end{align*} A very similar computation gives us that $$U_f(P) = a(b-a)+\frac{n+1}{n} \frac{(b-a)^2}{2}.$$ Letting $n$ approach $\infty$ we can see that we get the same limit namely \begin{align*} a(b-a) + \frac{(b-a)^2}{2} &= \frac{2ab - 2a^2 + b^2 + a^2 - 2ab}{2} = \frac{b^2 - a^2}{2} \end{align*}