Let $f_{n}:X\to Y$ be a sequence of bounded functions from one metric space $(X,d_{X})$ to another metric space $(Y,d_{Y})$. Suppose that $f_{n}$ converges uniformly to another function $f:X\to Y$. Suppose that $f$ is a bounded function; i.e., there exists a ball $B(y_{0},R)$ in $Y$ such that $f(x)\in B(y_{0},R)$ for all $x\in X$. Show that the sequence $f_{n}$ is uniformly bounded; i.e. there exists a ball $B(y_{0},R)$ in $Y$ such that $f_{n}(x)\in B(y_{0},R)$ for all $x\in X$ and all positive integers $n$.
MY ATTEMPT
Let $\varepsilon > 0$. Then there exists a natural number $N\geq 0$ such that for every $x\in X$ one has that \begin{align*} n\geq N \Rightarrow d_{Y}(f_{n}(x),f(x)) < \varepsilon \end{align*} Thus if we set $\varepsilon = 1$, there corresponds a natural number $N_{1}\geq 0$ such that \begin{align*} n\geq N_{1} \Rightarrow d_{Y}(f_{n}(x),f(x)) < 1 \end{align*}
Consequently, due to the triangle inequality, we have that \begin{align*} d_{Y}(f_{n}(x),y_{0}) \leq d_{Y}(f_{n}(x),f(x)) + d_{Y}(f(x),y_{0}) < 1 + R \end{align*} whenever $n\geq N_{1}$. Since the sequence is bounded, for $1\leq n\leq N_{1}$, we have that $f_{n}\in B(y_{0},r_{n})$.
Gathering these results, we conclude that $f_{n}(x)\in B(y_{0},r)$ for all $x\in X$ and $n\in N$, where $r = \max\{r_{1},r_{2},\ldots,r_{N_{1}},1+R\}$.
Yes, your proof is right. But you can also strengthen the claim: you don't need to assume that the uniform limit function $f$ is bounded, you can actually prove that this is the case.
To see this, choose $\epsilon = 1$. Then, there is an $N \in \Bbb{N}$ such that for all $n \geq N$, and all $x \in X$, we have \begin{align} d_Y(f_n(x), f(x)) < 1. \end{align} Let $r_N > 0$ be such that for all $x \in X$, $f_N(x) \in B(y_0, r_N)$. Then, the triangle inequality shows that for all $x \in X$, $f(x) \in B(y_0, 1+r_N)$.
So, in fact, $\{f, f_1, f_2, f_3, \dots\}$ are all uniformly bounded: if we set $r := \max\{r_1, \dots, r_N, 1+R_N\}$ then all these functions have image lying inside $B(y_0, r)$.