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To apply the Schröder-Bernstein theorem, we need injections in two directions. Given that the Schröder-Bernstein theorem doesn't require the axiom of choice (AC), there's a value in avoiding anything that requires AC (such as the theorem that existence of a surjection from $A$ to $B$ implies the existence of an injection from $B$ to $A$).

I'll construct the two injections explicitly; that way I can be sure to not accidentally use the AC. Note I'll use $\mathcal P(\mathbb R)$ to denote the power set of $\mathbb R$.

An injection $\mathcal P(\mathbb R)\to\mathbb R^{\mathbb R}$ can be given as
$$f(A) = x\mapsto\begin{cases}
1 & x\in A\\\\
0 & x\notin A
\end{cases}$$

For an injection $\mathbb R^{\mathbb R}\to \mathcal P(\mathbb R)$, we start with a bijection $\phi:\mathbb R\to (0,1)$ given by
$$\phi(x) = \exp(-\exp (x))$$

Then we define an injection $\iota:(0,1)\to \mathcal P(\mathbb N)$ by writing the number as dyadic fraction (choosing the period-0 representation in case of ambiguity), and then defining
$$\iota(x) = \{n\in\mathbb N: \text{the $n$-th digit is 1}\}$$

With those two ingredients, we now can build an injection $\alpha: \mathbb R^{\mathbb R}\to\mathcal P(\mathbb N\times\mathbb R)$ given by
$$\alpha(f) = \\{(n,x)\in\mathbb N\times\mathbb R:
n\in\iota\circ\phi\circ f\\}$$

Next, we define the *surjection* $\sigma:\mathbb R\to\mathbb N\times\mathbb R$ by
$$\sigma(x) = \begin{cases}
\Big(\Big|\lfloor x\rfloor\Big|,0\Big) & x\in\mathbb Z\\\\
\Big(\Big|\lfloor x\rfloor\Big|,\ln(-\ln(x-\lfloor x\rfloor))\Big) & \text{otherwise}
\end{cases}$$
Finally we define the function $g:\mathbb R^{\mathbb R}\to \mathcal P(\mathbb R)$ as
$$g(f) =\{x\in \mathbb R:\sigma(x)\in\alpha(f)\}$$

What remains is to prove that $g$ is indeed an injection. For this, assume $g(f_1) = g(f_2)$. Since $\sigma$ is surjective, this implies $\alpha(f_1) = \alpha(f_2)$, as otherwise there would be some $x$ such that $\sigma(x)$ is in exactly one of $\alpha(f_1)$ and $\alpha(f_2)$. But $\alpha$ in injective by construction, thus $f_1=f_2$.

Since we thus have, by explicit construction, both an injection $\mathcal P(\mathbb R)\to\mathbb R^{\mathbb R}$ and an injection $\mathbb R^{\mathbb R}\to \mathcal P(\mathbb R)$, we now can use Schröder-Bernstein to conclude that there exists a bijection between those sets.