Suppose $f:E\to E'$ is continuous, then if $N\subset E'$ is open, we know that $f^{-1}(N)\subset E$ must also be open. This is the open set condition for continuity. In the case of my course this was the definition of continuity. So, we basically want to show equivalence of the open and closed set conditions.
Suppose $f$ is continuous. Suppose some $C\subset E'$ is closed. Then, we know that $C^C$ is necessarily open. This implies that $f^{-1}(C^C)$ is also open. So, $(f^{-1}(C^C))^C$ must be closed, $(f^{-1}(C^C))^C=f^{-1}(C)$. Thus, $C$ closed implies $f^{-1}(C)$ closed for all continuous $f$.
Can you prove the converse of the statement?
A correction: the hypothesis for the other implication should read:
Now suppose that whenever $U$ is a closed set in $\mathbb{R}^n$, $f^{-1}(U)$ is a closed set in $\mathbb{R}^m$.
To prove this implication, you can show the contrapositive. Assume that $f$ is not continuous, so it is not continuous at some point $x_0 \in \mathbb{R}^m$. Then, we will find a closed set $U \subseteq \mathbb{R}^n$ such that $f^{-1}(U)$ is not a closed subset of $\mathbb{R}^m$.
Take a sequence $\{ x_n \}_{n \in \mathbb{N}}$ in $\mathbb{R}^m$ converging to $x_0$ such that $\{ f(x_n) \}_{n \in \mathbb{N}}$ does not converge to $f(x_0)$ in $\mathbb{R}^n$. This could mean one of two things: either $f(x_0)$ does not lie in the closure of $A = \{ f(x_n) : n \in \mathbb{N} \}$, or $f(x_0)$ does lie in the closure of $A$ but $f(x_n)$ does not converge to $f(x_0)$.
The first case is easy to dispose of: $f(x_0) \not\in cl(A) \implies x_0 \not\in f^{-1}(cl(A))$. But $x_0$ lies in the closure of $B = \{ x_n : n \in \mathbb{N} \}$, and $B \subseteq f^{-1}(cl(A))$, so we have a contradiction.
For the second case, if $f(x_n)$ does not converge to $f(x_0)$, then there exists a subsequence $\{ f(x_{n_k}) \}_{k \in \mathbb{N}}$ such that every term is "far away" from $f(x_0)$. That is, there exists $\epsilon_0 > 0$ such that $|f(x_{n_k}) - f(x_0)| \geq \epsilon_0$ for all $k \in \mathbb{N}$. Let $A_1 = \{ f(x_{n_k}) : k \in \mathbb{N} \}$. Then, $\{ x_{n_k} \}_{k \in \mathbb{N}}$ is a sequence in $\mathbb{R}^m$ converging to $x_0$, but $f(x_0)$ does not lie in the closure of the set $A_1$. This is the case dealt with previously, so again we are done.
Note that @karmalu's answer is incorrect by the following example. Consider the function $f : \mathbb{R} \to \mathbb{R}$ given by
$$
f(x) =
\begin{cases}
0, & x < 0;\\
1, & x \geq 0.
\end{cases}
$$
Choose the sequence $\{ x_n \}_{n \in \mathbb{N}}$ given by
$$
x_n = \frac{(-1)^n}{n}.
$$
Then, $x_n$ converges to the point $x_0 = 0$. Moreover, $f(x_0) = 1$ lies in the closure of $\{ f(x_n) \}_{n \in \mathbb{N}_{>i}}$ for each $i \in \mathbb{N}$, because
$$
f(x_n) =
\begin{cases}
0, & n \mathrm{\ odd};\\
1, & n \mathrm{\ even}.
\end{cases}
$$
Yet, $1$ is not the limit of $\{ f(x_{n}) \}_{n \in \mathbb{N}}$ because this is not a convergent sequence.
Best Answer
Let $\{f(c_n)\}$ be a sequence in $f(C)$ converging to some point $y$. We have to show that $y \in f(C)$. Let $K=\{y,f(c_1),f(c_2),\cdots \}$. Then $K$ is a compact set and hence $f^{-1}(K)$ is compact. Also, $\{c_n\} \subset f^{-1}(K)$. Hence there is a subseqeunce $c_{n_k}$ conveging to some limit $x$. Since $C$ is closed, $x \in C$. Since $f$ is continuous, $f(c_{n_k}) \to f(x)$. But then $y=f(x)$. Hence $y \in F(C)$ as required.