If Any continuous map from $A$ into $\mathbf{R}$ may be extended to a continuous map of all of $X$ into $\mathbf{R}$. Then $A$ is closed subset. TRUE/ FALSE:  Let $X$ be a normal space and $A$ be a subspace of $X$. If Any continuous map from $A$ into $\mathbf{R}$ may be extended to a continuous map of all of $X$ into $\mathbf{R}$. Then $A$ is closed subset.
My Attempt: I think if $X = \mathbf{R}^n$ or any norm linear space, then it is true. But what about arbitrary metric space or topological space.
 A: Recall that a subset $A$ of the topological space $X$ is called C-embedded (in $X$), if every continuous function $f: A \rightarrow \mathbb{R}$ can be continuously extended to $X$.
So, you are asking, if every C-embedded subset of a (normal) space is closed.
This is true, if $X$ is completely regular, and every singleton of $X$ is $G_\delta$ (no normality needed here). In particular, this is true in any metric space.
See proposition 2 in the paper by M. I. Garrido and J. A. Jaramillo, On the Converse Of Tietze-Urysohn’s Extension Theorem (which might also provide further information about your question, in fact, they directly address this question).
However, in general this is not true for normal spaces:

Consider $\beta \mathbb{N}$, the Stone-Cech compactification of the integers, which is compact, hence normal.
Let $x$ be a point of the remainder $\beta \mathbb{N} \setminus \mathbb{N}$ and $A := \beta \mathbb{N} \setminus \{x\}$. Then the Stone-Cech compactifaction of $A$ is $\beta \mathbb{N}$, hence $A$ is almost compact (i.e. $|\beta A \setminus A| \le 1$). Hence $A$ is C-embedded in any completely regular space, which it is a subspace of. In particular it is C-embedded in $\beta \mathbb{N}$ (see Gillman, Jerison, Rings of continuous functions, 6J). Of course, $A$ is not closed in $\beta \mathbb{N}$.
