Even on a compact interval instead of $\mathbb R$, this cannot be true...
Try $f_n$ defined on $[0,1]$ by
$$f_n(x)=nx^n(1-x),
$$
and
$$x_n=1-1/n.
$$
Then $x_n\to x$ with $x=$ $___$, $f_n\to f$ pointwise with $f=$ $___$, but $f_n(x_n)\to$ $___$ while $f(x)=$ $___$ hence the convergence cannot be uniform.
Note that the regularity of each $f_n$ is not an issue.
Another example: pick some nonzero continuous function $g$ such that $g(x)\to0$ when $|x|\to\infty$, say $g:x\mapsto\mathrm e^{-x^2}$, and define $f_n$ by $f_n(x)=g(n^2x-n)$ for every $x$.
Then $f_n\to f$ pointwise, where $f=$ $___$, but $f_n((n+x_0)/n^2)$ does not converge to $___$, for some well chosen $x_0$.
Fixing the problem statement
Here's a Wikipedia link with definitions of various types of equicontinuity. Normally, if you just say a sequence of functions is "equicontinuous" then it means they're pointwise equicontinuous. That assumption is not sufficient here and I'll provide a counterexample below.
However, the problem will work out fine if we instead assume the $\{f_n\}$ are uniformly equicontinuous, so I'm guessing that was the intended assumption. As to why it got miscommunicated - maybe your professor just defined "equicontinuous" to mean "uniformly equicontinuous", or maybe you made a typo in your question here, or maybe the professor made a typo in the problem statement.
Counterexample if we only assume pointwise equicontinuity
The problem as currently stated doesn't work. As a counterexample, let $f$ be some function that's continuous but not uniformly continuous; say $f(x) = \frac 1 x$ with $D = (0,1]$. Let $f_n = f$ for all $n$. Then the $f_n$ are continuous, and the sequence $\{f_n\}$ is equicontinuous (because all the functions are the same anyway), and clearly $f_n \to f$ pointwise. So the problem's assumptions are fulfilled but the claim doesn't hold.
Proof if we assume uniform equicontinuity
Choose some $\varepsilon > 0$. To show $f$ is uniformly continuous, we need to produce $\delta > 0$ such that if $a, b \in D$ with $|b-a|<\delta$ then $|f(b)-f(a)|<\varepsilon$.
To accomplish that using the $\frac{\varepsilon}{3}$ trick, the idea is to write
$$\begin{align}
|f(b) - f(a)| &= |f(b) - f_n(b) + f_n(b) - f_n(a) + f_n(a)-f(a)| \\
&\le |f(b) - f_n(b)| + |f_n(b) - f_n(a)| + |f_n(a)-f(a)|
\end{align}$$
where this inequality holds for any $n \in \mathbb{N}$. Now we aim to show that each of the three RHS terms can be made small.
For the 1st term, we know $f_n \to f$ pointwise, so there exists some $N_1 \in \mathbb{N}$ such that $|f(b) - f_n(b)| < \frac{\varepsilon}{3}$ for all $n \ge N_1$. Similarly, for the 3rd term we can find some $N_2 \in \mathbb{N}$ such that $|f_n(a) - f(a)| < \frac{\varepsilon}{3}$ for all $n \ge N_2$. That means if we pick $n \ge \max\{N_1, N_2\}$ then both of these terms will be $< \frac \varepsilon 3$.
To handle the 2nd term, we use the assumption that $\{f_n\}$ is a uniformly equicontinuous sequence. That means we can find $\delta > 0$ such that if $|x_2 - x_1| < \delta$ then $|f_n(x_2) - f_n(x_1)| < \frac{\varepsilon}{3}$ for all $n$. So as long as we require $|b-a| < \delta$, the 2nd term in our breakdown will be $< \frac{\varepsilon}{3}$.
Finally, we combine our estimates. For $n \ge \max\{N_1,N_2\}$, and for any $a,b \in D$ satisfying $|b-a| < \delta$, we have:
$$\begin{align}
|f(b) - f(a)| &\le |f(b) - f_n(b)| + |f_n(b) - f_n(a)| + |f_n(a)-f(a)| \\
&< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} \\
&= \varepsilon
\end{align}$$
which is what we needed to show for uniform continuity of $f$.
Best Answer
The statement cannot be proved, since it is false. Suppose, for instance, that you define, for each $n\in\mathbb N$,$$\begin{array}{rccc}f_n\colon&[0,1]&\longrightarrow&\mathbb R\\&x&\mapsto&x^n-x^{2n}.\end{array}$$Then each $f_n$ is continuous and $(f_n)_{n\in\mathbb N}$ converges pointwise to the null function (which is continuous too), but the convergence is not uniform:$$(\forall n\in\mathbb N):f_n\left(\sqrt[n]{\dfrac12}\right)=\frac14.$$