The standard proof (involving $\epsilon$-s and $\delta$-s) works even if $A=\mathbb R$. The most likely explanation is that $\subset$ in whatever book you are reading actually means $\subseteq$.
But let's assume we have only a proof of the following statement:
If $f_n: A\to\mathbb R$ converges to $f$ uniformly and $A\neq \mathbb R$ and $A\subset \mathbb R$, then $f$ is continuous.
We want to prove the statement:
If $f_n: \mathbb R\to\mathbb R$ converges to $f$ uniformly, then $f$ is continuous.
Let's take a sequence of functions $f_n$ such that $f_n:\mathbb R\to\mathbb R$ and $f_n$ converges to $f$ uniformly on $\mathbb R$.
Then, we can define $g_n:[-1,\infty)\to \mathbb R$ as $g_n(x)=f_n(x)$ for every $x\geq 0$. It is easy to show that $g_n$ converges to $g$, defined as $$g:[-1,\infty)\to \mathbb R\\
and, since $[0,\infty)\neq \mathbb R$, we conclude that $g$ is continuous.
Similarly, we can show that $h:(-\infty, 1]\to\mathbb R$, defined as $h(x)=f(x)$, is continuous.
Conclusion: $f$ is continuous.