Theorem: Let $(f_n)$ be a sequence of continuous functions. If $(f_n)$ converge uniformly to $f$, then $f$ is continuous.
I don't like the Rudin's approach, then I try to do my own way, but it seems to be wrong.
Proof: By definitions, we have:
$$\forall \varepsilon > 0$$ $$\exists n_1\in\mathbb{N}; n>n_1 \Rightarrow ||f_n(x)-f(x)||<\varepsilon/3$$ $$\exists n_2\in\mathbb{N}; n>n_2 \Rightarrow ||f_n(a)-f(a)||<\varepsilon/3$$ $$\exists \delta > 0; ||x-a||<\delta \Rightarrow ||f_n(x)-f_n(a)||<\frac{\epsilon}{3}$$
Put $n_0 = \max\{n_1,n_2\}$. Then we have
$$n>n_0 \text{ and } ||x-a||<\delta \Rightarrow ||f(x)-f_n(x)||+||f_n(a)-f(a)||+||f_n(x)-f_n(a)|| < \varepsilon$$
By triangular inequality:
$$||f(x)-f(a)||<\epsilon \text{ } \blacksquare$$
In your third step, $\delta$ is not necessarily the same for all $n$, i.e. $\delta = \delta_{n,\varepsilon,a}$ depends on $\varepsilon, a$ and $n$. So indeed, your proof is incorrect.
At a high-level, I'd encourage you to be more rigorous with the quantifiers (you miss the quantifiers relating to $n$ in the third part), and to write out explicitly the dependencies whenever possible ("$\exists \delta = \delta_{n,\varepsilon,a}$" for instance).
PS: last comment: in your first two statements, there should be no $x$. $f_n$, $f$ are functions; $f_n(x)$, $f(x)$ here are not .