Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and define a random variable $X: \mathcal{F}:\to \mathbb{R}.$ I want to show that the $\sigma$-field $\sigma(X)$ generated by $X$ is indeed a $\sigma$-field, and $\sigma(X) \subset \mathcal{F}.$
I'll first show that $\sigma(X) \subset \mathcal{F}.$ Choose an $A \in \sigma(X):$ $$ A = \lbrace X \in B \rbrace \in \sigma(X) $$ for some Borel set $B$. By definition of a random variable, $X$ is $\mathcal{F}$-measurable. This means $\lbrace X \in B \rbrace \in \mathcal{F}$ for all Borel sets $B \in \mathcal{B}(\mathbb{R})$. Thus $A \in \mathcal{F}$ which implies $\sigma(X) \subset \mathcal{F}.$
To show that $\sigma(X)$ is a $\sigma$-field, I have to show: (a) $\varnothing \in \sigma(X),$ (b) $A \in \sigma(X) \implies A^c \in \sigma(X)$, and (c) if $(A_n)_{n=1}^{\infty}$ is a sequence of sets in $\sigma(X)$ then $$ \bigcup_{n=1}^{\infty} A_n \in \sigma(X). $$ (a) Since $\mathcal{B}(\mathbb{R})$ is a $\sigma$-field, $\varnothing \in \mathcal{B}(\mathbb{R}).$ Choose $B = \varnothing.$ Then $$ \lbrace X \in \varnothing \rbrace = X^{-1}(\varnothing) = \varnothing \in \sigma(X). $$ (b) Let $A \in \sigma(X).$ Then $A = \lbrace X \in B \rbrace$ for some $B \in \mathcal{B}(\mathbb{R}).$ Since $\mathcal{B}(\mathbb{R})$ is a $\sigma$-field, $B^c \in \mathcal{B}(\mathbb{R})$. The set $A^c$ is the set $$ A^c = \lbrace \omega \in \Omega : X(\omega) \in B^c \rbrace = \{ X \in B^c \}, $$ and since $B^c \in \mathcal{B}(\mathbb{R}),$ $\lbrace X \in B^c \rbrace \in \sigma(X).$
(c) Suppose $(A_n)_{n = 1}^{\infty}$ is a sequence in $\sigma(X).$ Then for each $n$, $$ A_n = \lbrace X \in B_n \rbrace $$ for some $B_n \in \mathcal{B}(\mathbb{R}).$ Since $\mathcal{B}(\mathbb{R})$ is a $\sigma$-field and $B_n \in \mathcal{B}(\mathbb{R}),$ we must have that $\bigcup_{n=1}^{\infty} B_n \in \mathcal{B}(\mathbb{R}).$
I claim that $$ \bigcup_{n=1}^{\infty} A_n = \left\lbrace X \in \bigcup_{n=1}^{\infty} B_n \right\rbrace. $$ Pick $\omega \in \bigcup_{n=1}^{\infty} A_n.$ Then $\omega \in \lbrace X \in B_n \rbrace$ for some $n,$ implying $X(\omega) \in B_n$ for some $n$: $$ X(\omega) \in \bigcup_{n=1}^{\infty} B_n. $$ Thus $\omega \in \lbrace \omega \in \Omega: X(\omega) \in \bigcup_{n=1}^{\infty} B_n \rbrace = \lbrace X \in \bigcup_{n=1}^{\infty} B_n \rbrace$.
Now take $\omega \in \lbrace X \in \bigcup_{n=1}^{\infty} B_n \rbrace.$ Then $X(\omega) \in B_n$ for some $n$. So $\omega \in \lbrace X \in B_n \rbrace \subset \bigcup_{n=1}^{\infty} A_n.$
Therefore the equality claim is proven. Since $\bigcup_{n=1}^{\infty} B_n$ is a Borel set, $$ \bigcup_{n=1}^{\infty} A_n = \left\lbrace X \in \bigcup_{n=1}^{\infty} B_n \right\rbrace \in \sigma(X). $$