Let's first determine the sets of actions of the players.
An action of player 1 is simply a bid $x_1 \in \mathbb{R}_+$.
An action of player 2 is a function: $f_2: \mathbb{R}_+ \to \mathbb{R}_+$ that determines for every action $x_1$ of Player $1$ an action $x_2 = f_2(x_1) \in \mathbb{R}_+$. Let us denote by $F_2$ the set of all actions of player 2.
Let's now look at the payoffs:
$$
\begin{align*}
u_1(x_1, f_2) &= \left\{\begin{array}{ll} 500 - x_1 &\text{ if } x_1 > f_2(x_1)\\ \frac{500 - x_1}{2} &\text{ if } x_1 = f_2(x_1),\\
0 &\text{ if } x_1 < f_2(x_1)\end{array}\right.\\
u_2(x_1, f_2) &= \left\{\begin{array}{ll} 500 - f_2(x_1) &\text{ if } f_2(x_1) > x_1\\ \frac{500 - f_2(x_1)}{2} &\text{ if } x_1 = f_2(x_1),\\
0 &\text{ if } f_2(x_1) < x_1 \end{array}\right.
\end{align*}
$$
A strategy profile $(x_1^\ast, f_2^\ast)$ is a Nash equilibrium iff for all other $(x_1, f_2)$
$$
u_1(x_1^\ast, f_2^\ast) \ge u_1(x_1, f_2^\ast),\\
u_2(x_1^\ast, f_2^\ast) \ge u_1(x_1^\ast, f_2).
$$
Claim 1: If $(x_1^\ast, f_2^\ast)$ is a Nash equilibrium and if $x_1^\ast < 500$, then $f_2^\ast(x_1^\ast) \in (x_1^\ast, 500)$.
proof: Assume not, then either $f_2^\ast(x_1^\ast) \le x_1^\ast$ or $f_2^\ast(x_1^\ast) \ge 500$. Notice that $u_2(x_1^\ast, f_2^\ast) \le 0$. Now take the strategy $f_2$ where $f_2(x_1) = f_2^\ast(x_1)$ for all $x_1 \ne x_1^\ast$ and $f_2(x_1^\ast) = (500 +2 x_1^\ast)/3 > x_1^\ast$ . Then:
$$
> u_2(x_1^\ast, f_2) = 500 - \frac{500 + 2 x_1}{3} > \frac{500 - x_1^\ast}{2} \ge u_2(x_1^\ast, f_2^\ast),
> $$
Contradicting the definition of a Nash equilibrium.
Claim 2: If $(x_1^\ast, f_2^\ast)$ is a Nash equilibrium and if $x_1^\ast = 500$, then $f_2^\ast(x_1^\ast) \le 500$.
proof: Towards a contradiction, assume that $f_2^\ast(x_1^\ast) > 500$. Take the strategy $f_2$ where $f_2(x_1) = 500$ for all $x_1$. Then:
$$
> u_2(x_1^\ast, f_2^\ast) = 500 - f_2^\ast(x_1^\ast) < 0 = u_2(x_1^\ast, f_2)
> $$
This contradicts the definition of a Nash equilibrium.
Claim 3: If $(x_1^\ast, f_2^\ast)$ is a Nash equilibrium and if $x_1^\ast > 500$ then $f_2^\ast(x_1^\ast) < x_1^\ast$.
proof: Towards a contradiction, assume that $f_2^\ast(x_1^\ast) \ge x_1^\ast$. Take the strategy $f_2$ where $f_2(x_1) = 500$ for all $x_1$. Then:
$$
> u_2(x_1^\ast, f_2^\ast) \le \frac{500 - x_1^\ast}{2} < 0 = u_2(x_1^\ast, f_2).
> $$
This contradicts the definition of a Nash equilibrium
Claim 4: If $(x_1^\ast, f_2^\ast)$ is a Nash equilibrium, then $x_1^\ast \le 500$.
proof: Towards a contradiction, if $x_1^\ast > 500$ then we know from Claim 3 that that $f_2^\ast(x_1) < x_1^\ast$. As such:
$$
> u_1(x_1^\ast, f_2^\ast) = 500 - x_1^\ast < 0 \le u_1(500, f_2^\ast).
> $$
This contradicts the assumption of a Nash equilibrium.
Claim 5: There is no Nash equilibrium $(x_1^\ast, f_2^\ast)$ with $x_1^\ast < 500$.
proof: Assume that $(x_1^\ast, f_2^\ast)$ is a Nash equilibrium with $x_1^\ast < 500$. From Claim 1, we know that $f_2^\ast \in (x_1^\ast, 500)$. Take the strategy $f_2$ where $f_2(x_1) = \frac{x_1 + f_2^\ast(x_1)}{2}$. Notice that $f_2(x_1^\ast) > x_1^\ast$. Then:
$$
> u_2(x_1^\ast, f_2) = 500 - \frac{x_1 + f_2(x_1^\ast)}{2} > u_2(x_1^\ast, f_2^\ast),
> $$
This contradicts the assumption of a Nash equilibrium
Claim 6: There is no Nash equilibrium $(x_1^\ast, f_2^\ast)$ with $x_1^\ast > 500$.
proof: Towards a contradiction, assume that $(x_1^\ast, f_2^\ast)$ is a Nash equilibrium and $x_1^\ast > 500$. From Claim 3, we know that $f_2^\ast(x_1^\ast) < x_1^\ast$. Then:
$$
> u_1(x_1^\ast, f_2^\ast) = 500 - x_1^\ast < 0 = u_1(500, f_2^\ast).
> $$
Again a contradiction with the definition of a Nash equilibrium
The last two theorems show that if there is a Nash equilibrium $(x_1^\ast, f_2^\ast)$, then $x_1^\ast = 500$.
Claim 7 if $(x_1^\ast, f_2^\ast)$ is a Nash equilibrium then for all $x_1 < 500$, we have $f_2^\ast(x_1) > x_1$.
proof: Towards a contradiction, let $x_1 < 500$ and $f_2^\ast(x_1) \le x_1$ then:
$$
> u_1(x_1, f_2^\ast) \ge \frac{500 - x_1}{2} > 0 = u_1(x_1^\ast, f_2^\ast),
> $$
A contradiction with the definition of a Nash equilibrium
Claim 8 A strategy profile $(x_1^\ast, f_2^\ast)$ is a Nash equilibrium if and only if it of the following form:
$$
x_1^\ast = 500,\\
f_2^\ast(x_1) = \left\{\begin{array}{ll} \in [0,500] &\text{ if } x_1 = 500,\\
\in (x_1, + \infty) &\text{ if } x_1 < 500,\\
\in \mathbb{R}_+ &\text{ if } x_1 > 500. \end{array}\right.
$$
proof: $(\rightarrow)$ If $(x_1^\ast, f_2^\ast)$ is a Nash equilibrium, then $x_1^\ast = 500$ follows from Claim 5 and 6. Next, $f_2^\ast(500) = f_2^\ast(x_1^\ast) \in [0, 500]$ follows from Claim 2. $f_2(x_1) \in (x_1, + \infty)$ for $x_1 < 500$ follows from Claim 7. Finally, that $f_2^\ast(x) \in \mathbb{R}_+$ for $x > 500$ is obvious.
$(\leftarrow)$ Assume that $x_1^\ast = 500$ and that $f_2^\ast$ satisfies the restrictions in the claim. Notice that: $u_1(x_1^\ast, f_2^\ast) = 0$ and $u_2(x_1^\ast, f_2^\ast) = 0$.
Also, for all $f_2 \in F_2$, $u_2(x_1^\ast, f_2) \le 0$ and for all $x_1 \in \mathbb{R}_+$, we have $u_1(x_1) \le 0$. As such:
$$
> u_2(x_1^\ast, f_2) \le 0 = u_2(x_1^\ast, f_2^\ast),\\
> u_1(x_1, f_2^\ast) \le 0 = u_1(x_1^\ast, f_2^\ast).
> $$
So $(x_1^\ast, f_2^\ast)$ is a Nash equilibrium.
