Show that a multivariable function is (Riemann) integrable from definition This question is modified from Ex. 3.1 in Calculus on Manifolds by Spivak.

Let $f: R = [0,1] \times [0,1] \rightarrow \mathbb{R}$ be a bounded function defined by $f(x,y) := 0$ if $0 \leq x < 1/2$ and $f(x,y) := 1$ if $1/2 \leq x \leq 1$. Prove, using the definition, that $f$ is integrable and ${\displaystyle \int f}$ over $R = 1/2$.

My approach is to take a suitable partition $\mathcal{P}$ of $R$ such that $\inf U(f,\mathcal{P}) \leq \sup L(f,\mathcal{P})$. Then since $\inf U(f,\mathcal{P}) \geq \sup L(f,\mathcal{P})$ is always true, we have $\inf U(f,\mathcal{P}) = \sup L(f,\mathcal{P})$. However, I cannot find such partition $\mathcal{P}$. The partition $ \{[0,1/2] \times [0,1], [1/2,1] \times [0,1]\} $ seems not working. Is my approach correct?
Please note that I cannot use the Riemann Condition as the question requires a proof by definition.

Taking the partition $P=\{[0,1/2-\epsilon]\times [0,1], [1/2-\epsilon,1/2]\times [0,1], [1/2,1]\times[0,1]\}$ for any $\epsilon > 0$, we have $L(f,P) = 1/2$ and $U(f,P) = 1/2+\epsilon$.  These values for the upper and lower sums are attained because
$$\inf \{ f(x,y) : 1/2-\epsilon \leqslant x \leqslant 1/2, 0\leqslant y \leqslant 1\}= 0, \\\sup \{ f(x,y) : 1/2-\epsilon \leqslant x \leqslant 1/2, 0\leqslant y \leqslant 1\}= 1$$
Thus,
$$1\tag{*}/2 \leqslant \sup_P L(f,P) \leqslant \inf_P U(f,P) \leqslant 1/2+ \epsilon$$
If it were true that $\inf_P U(P,f) > 1/2$, then we could choose an $\epsilon_0$ such that $\inf_PU(f,P) > 1/2 + \epsilon_0$ which contradicts (*).
Therefore,
$$\sup_P L(f,P) = \inf_P U(f,P) = 1/2$$
