Let $ A = \{ f \in \mathbb{R}^\mathbb{R} \mid f \text{ is periodic }\} $. I want to prove that its cardinality is $ \aleph_1 $. Can you verify my proof?
First, $ A \subseteq \mathbb{R}^\mathbb{R} \Rightarrow |A| \le |\mathbb{R}^\mathbb{R}| = \aleph^\aleph = \aleph_1 $.
Let $ B = [0, 1)^{[0,1)} $.
We will define a function $ m: B \to A $.
Let $ g \in B $. We shall create a new function $ h $ in the following way:
Let $ x $ be a real number. Therefore: $ 0 \le x - \lfloor x \rfloor < 1 $.
Define $ h(x) = g(x - \lfloor x \rfloor) $.
We will show $ h $ is periodic. Let $ x $ be a real number. Therefore $ x - \lfloor x \rfloor = x + 1 - \lfloor x + 1 \rfloor $ which means $ h(x) = h(x + 1) $. Therefore, $ h $ is periodic.
Define $ m(g) = h $.
We will show $ m $ is injective. Let $ g_1, g_2 \in B $ and assume $ g_1 ≠ g_2 $. Therefore, there exists $ x $ such that $ g_1(x) ≠ g_2(x) $. $ x = x - \lfloor x \rfloor $ and therefore $ m(g_1)(x) ≠ m(g_2)(x) \Rightarrow m(g_1) ≠ m(g_2) $. Therefore, $ m $ is injective. Hence, $ \aleph_1 \le |A| $ because $ |B| = \aleph_1 $.
Therefore, by CSB, we get that $ |A| = \aleph_1 $.
I can verify that the logic of your proof is sound. My only concern is that your notation for cardinality does not seem to be standard. $\aleph_1$ normally denotes the smallest uncountable cardinal, so $|\mathbb{R}^\mathbb{R}|=2^{|\mathbb{R}|}>|\mathbb{R}|\ge\aleph_1$. ($|\mathbb{R}|=\aleph_1$ is the continuum hypothesis, which is independent of ZFC.) The standard notation would be $|\mathbb{R}^\mathbb{R}|=2^{\mathfrak{c}}$, where $\mathfrak{c}=2^{\aleph_0}$ is the cardinality of $\mathbb{R}$. Alternatively, $2^{\mathfrak{c}}=\beth_2$.