Let $(f_{n})_{n=1}^{\infty}$ be a sequence of functions from one metric space $(X,d_{X})$ to another $(Y,d_{Y})$, and suppose that this sequence converges uniformly to another function $f:X\to Y$. If the functions $f_{n}$ are bounded on $X$ for each $n$, then the limiting function $f$ is also bounded on $X$.
My solution
If $f_{n}\to f$ uniformly, for every $\varepsilon > 0$ there is a natural number $N\geq 1$ such that for every $x\in X$ we have that \begin{align*} n\geq N \Rightarrow d_{Y}(f_{n}(x),f(x)) < \varepsilon \end{align*}
Since each $f_{n}$ is bounded, we have that $f_{n}(X)\subseteq B(y_{0},r_{n})$.
But I do not know how to proceed from here. Could someone help me to solve this?
Edit
As suggested in the comments, we have that \begin{align*} d_{Y}(y_{0},f(x)) \leq d_{Y}(y_{0},f_{n}(x)) + d_{Y}(f_{n}(x),f(x)) \leq r_{n} + \varepsilon \end{align*}
Thus if we set $\varepsilon = 1$, there is a natural number $N_{1}\geq 1$ such that for every $x\in X$ one has that \begin{align*} d_{Y}(y_{0},f(x)) \leq r_{N_{1}} + 1 \end{align*} That is to say, $f(x)\in B(y_{0},r_{N_{1}}+1)$ for every $x\in X$, and we are done.