The theorem says :
If each $v_n(x)$ is continuous and $v_n(x)\rightarrow v_*(x)$ converges uniformly (as $n \rightarrow \infty$), then $v_*(x)$ is also a continuous function.
I tried to deduce the following statement, if $v_*(x)$ is continuous and $v_n(x)$ is as well continuous then the function sequence must converge uniformly.
But opening the book I read:
Warning: A sequence of continuous functions can converge nonuniformly to a continuous function.
How is that? Isn't this a contradiction to the implication of the theorem?
Your deduced statement is not true: if $v(x) = 0$ and $v_n (x) = \begin{cases} 0, & x \le n \\ x-n, & n < x \le n+1 \\ 1, & n+1 < x \end{cases}$, then $v_n \to v$, they are all continuous, defined on $\Bbb R$, but $ \sup | v_n(x) - v(x)| = 1 \not\to 0$, which shows that the convergence is not uniform.