Quick one: $\textbf{f}$ continuous , implies $F_a = \{ \textbf{x} \in E : \textbf{f}(\textbf{x}) = \textbf{f} (\textbf{a}) \}$ closed. $\textbf{Claim:}$ If we have a continuous mapping $\textbf{f}$ of an open set $E \subset R^n$ into $R^m$, then for any fixed $\textbf{a} \in E$, the set $F_a = \{ \textbf{x} \in E : \textbf{f}(\textbf{x}) = \textbf{f} (\textbf{a}) \}$ is closed.
$\textbf{Attempt to prove claim:}$ Consider any point $\textbf{x} \in F_a$. Now consider an an arbitrary neighborhood of this point $N_{r} (\textbf{x}) = \{\textbf{y} \in F_a : d(\textbf{x},\textbf{y}) < r \}$. For the set $F_a$ to be closed, $N_{r}(\textbf{x})$ must contain a point $\textbf{y} \neq \textbf{x}$  such that $\textbf{y} \in F_a$ (which is to say $\textbf{x}$ a limit point of $F_a$). By continuity of $\textbf{f}$, for all $\epsilon > 0$ there exists a $\delta > 0$ such that $d(\textbf{f}(\textbf{x}),\textbf{f}(\textbf{y})) < \epsilon$ whenever $d(\textbf{x},\textbf{y}) < \delta$. 
Now I am a bit stuck; I don't see how this line of reasoning could lead to $\textbf{f} (\textbf{y}) = \textbf{f}(\textbf{x})$. Is the claim false, and if so, can we add a ``reasonable'' hypothesis (such as requiring $\textbf{f}$ to be differentiable) to make the modified claim true?
 A: Note that $F_a$ is just the pre-image under $f$ of $\{f(a)\}$. Since $\{f(a)\}$ is closed and $f$ is continuous, its preimage is closed as well.
A: Below are two alternative proofs. Let $c \triangleq f(a)$.
Proof 1: To prove that a set is closed, you need to show that a limit point of a set is contained withing the set. That is to say, let $x^*$ be a limit point of the set $S = \{x\in E :f(x) = c \}$. First, since $x^{*}$ is a limit point, there is a sequence $\{x_n\}_{n=1}^{\infty}$ in the set such that $x_n \to x^*$. Now, you can prove that a function $f$ is continuous at a point $x^*$ iff for every sequence $x_n\to x^*$, $f(x_n) \to f(x^*)$. Since $f(x_n) = c$, $f(x^*) = c$, proving that $x^* \in S$.
Proof 2: Alternatively, you can prove that the complement of $S$, $S^c$ is an open set. Note that
$$
S^c = \{x \in E : f(x) \neq c\} = \underbrace{\{x \in E : f(x) > c\}}_{\triangleq S_1} \bigcup \underbrace{\{x \in E : f(x) < c\}}_{\triangleq S_2}.
$$
We will now prove that $S_1$ and $S_2$ are both open sets, proving that so do their union, hence $S^c$. 
Let $x \in S_2$. $f(x) <c$. Now, take an $\epsilon>0$ such that $\epsilon < c- f(x)$. For this $\epsilon>0$, take a $\delta>0$ (existence of such a $\delta$ is justified by continuity) such that
$$
d(x,y)<\delta \implies |f(x)-f(y)|<\epsilon,
$$
hence,
$$
f(y)<f(x)+\epsilon<c \implies y \in S_2
$$
for every $y \in B(x,\delta)$. Hence, for every $x \in S_2$, $\exists \delta > 0$ such that, $B(x,\delta) \subset S_2$, proving that $S_2$ is open. Now, you can do the same thing for $S_1$ and conclude.
