Theorem: Let $(f_n)$ be a sequence of functions on $I \subseteq \mathbb{R}$. Then $(f_n)$ pointwise convergent iff pointwise cauchy.
Here, I only prove "$\Longleftarrow$" since the converse is very straightforward.
Proof attempt: Suppose $\left(f_n \right)$ cauchy, then $\forall \epsilon >0 \: \forall x \in I \: \exists N_o \in \mathbb{N}: \forall m,n \in \mathbb{N}$ $$n,m \geq N_0 \implies \mid f_n(x)-f_m(x) \mid < \frac{\epsilon}{3} $$
Let $M$ be an upper bound of $f_{N_0}(x)$. Then $$\mid f_n(x)-f_{N_0}(x) \mid < \frac{\epsilon}{3} \implies \mid f_n(x) \mid < \frac{\epsilon}{3}+M$$ and $n$ is arbitrary, so $(f_n)$ is bounded. Now by Bolzano-Weierstrass theorem, $(f_n)$ has a convergent subsequence. Let $(f_{n_{k}})$ be a such sequence and $(f_{n_{k}}) \to f$. Then $\exists N_1 \in \mathbb{N}:$ $$n_k \geq N_1 \implies \mid f_{n_{k}}(x)-f(x) \mid<\frac{\epsilon}{3} $$ Combining the terms yields $$\mid f_{n_{k}}(x)-f(x)+f_n(x)-f_{N_0}(x) \mid \leq \mid f_{n_{k}}(x)-f(x) \mid + \mid f_n(x)-f_{N_0}(x) \mid <\frac{2\epsilon }{3}$$ Now, let $n_k \geq N_0$. Then $$\mid f_n(x)-f(x)\mid <\frac{2\epsilon }{3}+\mid f_{N_0}(x)-f_{n_k}(x)\mid $$ But since $n_k \geq N_0$, $\mid f_{N_0}(x)-f_{n_k}(x)\mid < \frac{\epsilon}{3}$ so $\mid f_n(x)-f(x)\mid <\epsilon$. And we can conclude that $(f_n) \to f$ $\square$
It seems fine to me except that you should take $n_k \geq N_1$ in the last part of the argument instead.
The result also follows immediately from the fact that a sequence of real number $a_n$ is convergent if and only if it is Cauchy.