Let $x = y$ in (2), we have $f\left(x^{2}\right) \geq f^{2}(x) \geq 0$. This implies that $f(x) \geq 0$ for all $x \geq 0$.

Let $x = 0$ in (1), we have $f(1) \geq f(0) + 1 \geq 1$.

Let $x = y = 1$ in (1), we have $f(1) \geq f^{2}(1)$, i.e., $0 \leq f(1) \leq 1$. These inequalities imply that $f(1) = 1$, and hence $f(0) = 0$.

From (1) and (2), by induction, we have
$$f(x+n) \geq f(x) + n \tag{*}$$
for all $x \in \mathbb{R}$, $n \in \mathbb{N}$, and
$$f\left(x^{n}\right) \geq f^{n}(x) \tag{**}$$
for all $x \geq 0$ and $n \in \mathbb{N}^{*}$.

Using $(*)$, for each $x > 0$, we have
$$f(x) = f(\{x\} + \lfloor x \rfloor) \geq f(\{x\}) + \lfloor x \rfloor \geq \lfloor x \rfloor > x - 1.$$
From $(**)$ and (2), for each $x > 1$ and $n \in \mathbb{N}^{*}$, we have
$$\left(x^{n} - 1\right) f^{n}\left(\frac{1}{x}\right) \leq f\left(x^{n}\right) f\left(\frac{1}{x^{n}}\right) \leq f(1) = 1.$$
Hence,
$$f\left(\frac{1}{x}\right) \leq \frac{1}{\sqrt[n]{x^{n} - 1}}$$
for all $x > 1$, $n \in \mathbb{N}^{*}$. Fix $x > 1$ and let $n \rightarrow +\infty$, we obtain $f\left(\frac{1}{x}\right) \leq \frac{1}{x}$, or (note that $f(0) = 0$ and $f(1) = 1$),
$$f(x) \leq x$$
for $x \in [0, 1]$.

On the other hand, we can prove by induction from (1) that
$$f(x-n) \leq f(x) - n$$
for $n \in \mathbb{N}$. For each $x < 0$, we have $\lfloor x \rfloor < 0$ so
$$f(x) = f(\{x\} + \lfloor x \rfloor) \leq f(\{x\}) + \lfloor x \rfloor \leq \{x\} + \lfloor x \rfloor = x.$$
In (2), set $x = y = t$ with $t \in [-1, 0]$ then
$$f^{2}(t) \leq f\left(t^{2}\right) \leq t^{2}$$
This implies that $t \leq f(t) \leq -t$. Hence, $f(t) = t$ for all $t \in [-1, 0]$.

Now, for each $x < -1$, we have
$$\frac{1}{x^{2}} f^{2}(x) = f^{2}(x) f^{2}\left(\frac{1}{x}\right) \leq f\left(x^{2}\right) f\left(\frac{1}{x^{2}}\right) \leq f(1) = 1$$
so $f^{2}(x) \leq x^{2}$, implying that $x \leq f(x) \leq -x$. So $f(x) = x$ for all $x \leq 0$.

Let $y = -1$ in (2) and consider $x > 0$, we have
$$-x = f(-x) \geq f(x) f(-1) = -f(x)$$
implying $f(x) \geq x$ for all $x > 0$. Hence, we have $f(x) = x$ for all $x \in (0, 1)$.

Finally, suppose that $x > 1$, we have
$$\frac{1}{x} f(x) = f(x) f\left(\frac{1}{x}\right) \leq f(1) = 1$$
so $f(x) \leq x$. Since $f(x) \geq x$, we have $f(x) = x$ for all $x > 1$.

Inclusion, $f(x) = x$ for every $x$ is the only function satisfying given conditions.