Suppose that $(f^{(n)})_{n=1}^{\infty}$ is 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 $f:X\to Y$. Let $x_{0}$ be a point of $X$. If the functions $f^{(n)}$ are continuous at $x_{0}$ for each $n$, then the limit function $f$ is also continuous at $x_{0}$
My solution (edit)
According to the comments, we have the following solution.
Let us start with the definition of uniform convergence.
For every $\varepsilon/3 > 0$ there is a natural number $N\geq 1$ such that for every $x\in X$ we have that $d_{Y}(f^{(n)}(x),f(x)) < \varepsilon/3$ for every $n\geq N$.
Thus, if we fix $n = N$, since each $f^{(n)}$ is continuous at $x_{0}$, we conclude that for every $\varepsilon/3 > 0$ there is a $\delta > 0$ such that for every $x\in X$ the relation holds
\begin{align*}
d_{X}(x,x_{0}) < \delta \Rightarrow d_{Y}(f^{(N)}(x),f^{(N)}(x_{0})) < \varepsilon/3
\end{align*}
Hence for every $\varepsilon > 0$ there is a $\delta > 0$ such that for every $x\in X$ we have that
\begin{align*}
d_{Y}(f(x),f(x_{0})) \leq d_{Y}(f(x),f^{(N)}(x)) + d_{Y}(f^{(N)}(x),f^{(N)}(x_{0})) + d_{Y}(f^{(N)}(x_{0}),f(x_{0})) < \varepsilon
\end{align*}
whenever $d_{X}(x,x_{0}) < \delta$. Therefore $f$ is continuous at $x_{0}$, just as it has been claimed.
Best Answer
Let $\epsilon >0$, then there exists $N\in \mathbb N $ and $\delta >0$ such that $d(x,x_0)<\delta \Rightarrow d(f_N(x),f_N(x_0))< \epsilon /3$ and $d(f_N(y),f(y))< \epsilon /3$ for any chosen $y$, therefore
$$ \begin{align*} d(f(x),f(x_0))&\leqslant d(f(x),f_N(x))+d(f_N(x),f(x_0))\\ &\leqslant d(f(x),f_N(x))+d(f_N(x),f_N(x_0))+d(f_N(x_0),f(x_0))\\ &< \epsilon \end{align*} $$ that is, we had shown that $d(x,x_0)< \delta \Rightarrow d(f(x),f(x_0))< \epsilon $, so $f$ is continuous at $x_0$.