I'm trying to show the following.
Let $f_k:A\rightarrow \mathbb R$ be unbounded for each $k \in \mathbb N$ and $f_k$ converge to $f$ uniformly. Then $f$ is also unbounded.
My proof:
Supposed $f_k:A\rightarrow \mathbb R$ is unbounded for each $k \in \mathbb N$ and $f_k$ converge to $f$ uniformly. Since for each $k\in\mathbb N$ $f_k$ is unbounded, for all $M \in \mathbb R$ there is some $x_0\in A$ such that $\left|f_k(x_0)\right|>M+1 \Leftrightarrow \left|f_k(x_0)\right|-1>M$. Also, since $f_k\rightarrow f$ uniformly, then for all $\varepsilon>0$ there is some $N\in\mathbb N$ such that $\left|f(x)-f_k(x)\right|<\varepsilon$ for $k\ge N$ and for all $x \in A$.
We also know from the reverse triangle inequality that
$\left|\left|f(x)\right| - \left|f_k(x)\right|\right| \le \left|f(x)-f_k(x)\right|$.
So if we let $\varepsilon = 1$, then for some $N\in \mathbb N$ we have
$\left|\left|f(x)\right| - \left|f_N(x)\right|\right| \le \left|f(x)-f_N(x)\right|<1$
$\Rightarrow \left|\left|f(x)\right| - \left|f_N(x)\right|\right|<1$
$\Rightarrow -1< \left|f(x)\right| - \left|f_N(x)\right| <1$
$\Rightarrow \left|f_N(x)\right|-1 <\left|f(x)\right| $.
Evaluating this at the $x_0$ from earlier, we get
$\Rightarrow \left|f_N(x_0)\right|-1 <\left|f(x_0)\right| $
$\Rightarrow M<\left| f(x_0) \right|$.
Are there any flaws in my proof?