$X \subset \mathbb{R}^m$. Let $\phi:x \to \mathbb{R}^n$ be bounded. $\phi$ is continuous $\iff$ its graph is closed. 
$X \subset \mathbb{R}^m$. Let $\phi:x \to \mathbb{R}^n$ be bounded. Then $\phi$ is continuous $\iff$ its graph is closed.

I was asked to prove this, but I believe it's false.
Let $Gra(\phi) = \{(x, \phi(x))|x \in X\}$ the Graph of $\phi$.
Suppose X is not closed. Then exists $a \in \bar{X}-{X}$ (being $\bar{X}$ the closure of X).
Therefore exists $x_n \in X$ s.t. $x_n \to a$. The sequence $\phi(x_n)$ has a convergent subsequence $\phi(x_{k_n})$ since it's bounded. Let $\lim \phi(x_{k_n})=b$
Then $(a,b) \in \overline{Gra(\phi)}$  (the closure of $Gra(\phi)$) but $(a,b)\not \in Gra(\phi) \implies Gra(\phi)$ is not closed.
Is this correct?
 A: Of course, when $X$ itself is not closed in $\mathbb{R}^m$, the graph cannot be closed. We henceforth assume that $X$ is closed.
Let $\mathcal{C}(X)$ denote the set of convergent sequences $\mathbf{x}=(x_n)_{n\in\mathbb{N}}\subset X$. For any such sequence we will denote its limit by $x$. Notice that $x\in X$ by the assumption that $X$ is closed.
On the one hand, we have:
\begin{align}
\phi \text{ is continuous}&\iff \phi \text{ is sequentially continuous}\\
&\iff \forall \mathbf{x}\in\mathcal{C}(X), \,\phi(x)=\lim_{n\to\infty}\phi(x_n)\tag{1}\end{align}
On the other, we have
\begin{align}
Gra(\phi) \text{ is closed}&\iff Gra(\phi) \text{ is sequentially closed}\\
&\iff \forall \text{ convergent } \big(x_n,\phi(x_n)\big),\, \lim_{n\to\infty}\big(x_n,\phi(x_n)\big) \in Gra(\phi)\tag{2}
\end{align}
It is clear that $(1)\implies (2)$.
Indeed, if $\big(x_n,\phi(x_n)\big)$ is convergent, then $\mathbf{x}=(x_n)$ is convergent and hence by $(1)$ so too is $\big(\phi(x_n)\big)$,
with $\lim_{n\to\infty}\phi(x_n)=\phi(x)$.
Hence,
$$\lim_{n\to\infty}\big(x_n,\phi(x_n)\big)=\big(x,\phi(x)\big),$$
which belongs to $Gra(\phi)$ by definition.

We will show that when $\phi$ is bounded, $(2)\implies(1)$, which conludes the proof.
Suppose $\phi$ is bounded and let $\mathbf{x}\in\mathcal{C}(X)$.
Since $\phi$ is bounded and $(x_n)$ is convergent (and hence bounded), we have that $\big(\phi(x_n)\big)$ is a bounded sequence.
By the Bolzano-Weierstrass Theorem, there is a convergent subsequence $\big(\phi(x_{n_k})\big)$ of $\big(\phi(x_n)\big)$.
Of course, $(x_{n_k})$ it itself convergent and converges to $x$, the same limit of $(x_n)$.
It follows that $\big(x_{n_k},\phi(x_{n_k})\big)$ is convergent, and by $(2)$ we have that $\lim_{k\to\infty}\big(x_{n_k},\phi(x_{n_k})\big)=\big(x,\phi(x)\big)$.
In particular, $\lim_{k\to\infty}\phi(x_{n_k})=\phi(x)$.
It then suffices to show that $\big(\phi(x_n)\big)$ is convergent;
in this case, its limit must coincide with that of $\big(x_{n_k},\phi(x_{n_k})\big)$, that is, must equal $\phi(x)$. We show this by contradiction.
Indeed, if that were not the case, then there would be some $\epsilon>0$ and a subsequence $(x_{m_k})$ of $(x_n)$ with $d\big(\phi(x_{m_k}),\phi(x)\big)\geq \epsilon$ for all $k$.
Now, $\big(\phi(x_{m_k})\big)$ is of course bounded, so by the Bolzano-Weierstrass Theorem it must have a convergent subsequence, say $\left(\phi\left(x_{m_{k_j}}\right)\right)$.
By construction, we have that $$y=\lim_{j\to\infty}\phi\left(x_{m_{k_j}}\right)\neq \phi(x)$$
But $\left(x_{m_{k_j}}\right)$ is a subsequence of $(x_n)$, and hence converges to $x$.
It follows that $$\lim_{j\to\infty}\left(x_{m_{k_j}},\phi\left(x_{m_{k_j}}\right)\right)=(x,y)$$
and hence does not belong to $Gra(\phi)$, in contradiction with $(2)$. $\square$.


Notice that in the proof above we didn't actually need the hypothesis that $\phi$ be bounded, but rather that $\phi$ 'preserves boundedness', ie, $\phi$ takes bounded sets to bounded sets.

A: This is wrong (for unbounded function). take $\phi : [0, \infty] \rightarrow R$ with $\phi(x) = \frac{1}{x}$ for all $x \in (0, \infty)$ and $\phi (0)=0$. its graph is closed but $\phi$ is not continuous!
EDIT: The above example is for unbounded function on limited domain (before editing the original question). If we assume function is bounded then the claim  is correct.
Proof: For Left to Right: the function $F(x) = (x, \phi(x))$  is continuous (now to show graph is closed in $X \times R^n$, use the sequential definition of continuity of $F$).
For Right to Left: similar to your argument, take $x_n \rightarrow x \in X$. want to show that $\phi(x_n) \rightarrow \phi(x),$ if actually $\phi(x_n) \nrightarrow \phi(x)$ then (since $\phi$  is bounded)  $\phi (x_n)$ has a convergent subsequence, say $\phi (x_{n_{k}}) \rightarrow y \neq \phi(x)$, i.e., $$(x_{n_{k}},  \phi (x_{n_{k}}) ) \rightarrow (x,y)\notin \text{Graph}.$$
Which is contradicting with Graph being Closed. 
A: 
$\iff$ its graph is closed.

That's ambiguous. Closed in what space? To be precise, usually the graph is considered closed in $X\times\mathbb{R}^n$ (as opposed to $\mathbb{R}^m\times\mathbb{R}^n$ which you consider in your counterexample).
That's reasonable because your argument is correct if we consider whole $\mathbb{R}^m\times\mathbb{R}^n$. What you've shown is that if $X$ is not closed in $\mathbb{R}^m$ then no graph can be closed in $\mathbb{R}^m\times\mathbb{R}^n$ for bounded functions. IMO this makes $\mathbb{R}^m\times\mathbb{R}^n$ uninteresting (in this context).

Then $(a,b)\in \overline{Gra(\phi)}$

That's were you are wrong (at least under the proper assumption that we are considering $X\times\mathbb{R}^n$ as the universe). That point doesn't even belong to $X\times\mathbb{R}^n$ (because $a\not\in X$) so how can it belong to some subset? Your domain is $X$, not $\mathbb{R}^m$. Remember that the graph is defined as a subspace of domain times codomain and so the closure is also considered in $X\times\mathbb{R}^n$.

So here's the real proof. First of all I assume that by "limited" you actually mean bounded.
If $\phi$ is bounded, then that means that the image of $\phi$ is bounded. Thus
$$\mbox{im}(\phi)\subseteq C\subset\mathbb{R}^n$$
where $C$ is a compact subset of $\mathbb{R}^n$ (e.g. some closed ball). Now define
$$\phi':X\to C$$
$$\phi'(x)=\phi(x)$$
Note that $\phi$ is continous if and only if $\phi'$ is. Also graphs of $\phi$ and $\phi'$ are equal as sets.
On the other hand $\phi'$ is continous if and only if its graph is closed in $X\times C$ by the closed graph theorem (which applies since $C$ is compact). But since $X\times C$ is a closed subset of $X\times\mathbb{R}^n$ then the graph of $\phi'$ is closed in $X\times C$ if and only if it is closed in $X\times\mathbb{R}^n$.
All in all the following statements are equivalent:


*

*$\phi$ is continous

*$\phi'$ is continous

*graph of $\phi'$ is closed in $X\times C$

*graph of $\phi'$ is closed in $X\times\mathbb{R}^n$

*graph of $\phi$ is closed in $X\times\mathbb{R}^n$



This can be generalized as follows:

Let $X, Y$ be topological spaces with $Y$ Hausdorff. Assume that $f:X\to Y$ is a function such that $\overline{\mbox{im}(f)}$ is compact. Then $f$ is continous if and only if the graph of $f$ is closed in $X\times Y$.

