What does continuity of inclusion means? If $A,B,C$ are three spaces such that $A\subset B\subset C$ and $A$ is dense in $C$. Now my teacher said that the inclusion between three spaces are continuous and so you can directly say that $B$ is dense in $C$ . 
When i asked  him why can't you directly conclude this, he said that the topologies are different and so you need the condition that the inclusions are continuous. I didn't understand what he meant by that?
 A: Edit: Just to make things more precise, let's say that we're dealing with spaces $\langle A,\tau_A\rangle,$ $\langle B,\tau_B\rangle,$ and $\langle C,\tau_C\rangle,$ where $A\subseteq B\subseteq C$. In that case, you are quite correct that there is no need to know anything about the continuity of the inclusions. In fact, we don't even need to know anything at all about the topologies $\tau_A$ or $\tau_B$.
Given a topological space $\langle X,\tau\rangle$ and any $S\subseteq X$, observe that there is at least one $V\in\tau$ such that $S\cap V=\emptyset$--for example, $V=\emptyset$--and we denote/define the closure of $S$ in $\langle X,\tau\rangle$ by $$\text{cl}(S,X,\tau):=\bigcap\{X\smallsetminus V:V\in\tau,S\cap V=\emptyset\}.$$ This has the following properties (that you should verify):

(i) $S\subseteq\text{cl}(S,X,\tau)\subseteq X.$
(ii) $\text{cl}(S,X,\tau)$ is closed in $\langle X,\tau\rangle.$
(iii) If $F\subseteq X$ with $F$ closed in $\langle X,\tau\rangle$ and $S\subseteq F$, then $\text{cl}(S,X,\tau)\subseteq F$.

Putting (i)-(iii) together, it means that $\text{cl}(S,X,\tau)$ is the smallest subset of $X$ containing $S$ that is closed in $\langle X,\tau\rangle.$ Another property that you should verify is that $S\subseteq X$ is dense in $\langle X,\tau\rangle$ if and only if $\text{cl}(S,X,\tau)=X$.
By way of contradiction, suppose that $B$ is not dense in $\langle C,\tau_C\rangle,$ so that there is some $c\in C$ such that $c\notin\text{cl}(B,C,\tau_C),$ meaning by definition of $\text{cl}(B,C,\tau_C)$ that there is thus some $V_0\in\tau_C$ with $B\cap V_0=\emptyset$, such that $c\notin C\smallsetminus V_0$. Now, since $B\cap V_0=\emptyset$ and $A\subseteq B$, then $A\cap V_0=\emptyset$. But then by definition of $\text{cl}(A,C,\tau_C),$ we have that $$\text{cl}(A,C,\tau_C)\subseteq C\smallsetminus V_0,$$ so $c\notin\text{cl}(A,C,\tau_C),$ contradicting our assumption that $A$ is dense in $\langle C,\tau_C\rangle.$ Another way to approach it is to show that if $A\subseteq B\subseteq C$, then $\text{cl}(A,C,\tau_C)\subseteq\text{cl}(B,C,\tau_C),$ so if $A$ is dense in $\langle C,\tau_C\rangle,$ then $$C=\text{cl}(A,C,\tau_C)\subseteq\text{cl}(B,C,\tau_C)\subseteq C,$$ so $\text{cl}(B,C,\tau_C)=C$, as desired.

It is possible that, instead, your teacher meant that $A,B,C$ are spaces and that there exist injections $\iota_1:A\hookrightarrow B$ and $\iota_2:B\hookrightarrow C$, with $A$ densely imbedded into $C$ by $\iota_2\circ\iota_1$. That means that $\iota_2\circ\iota_1(A)$ dense in $C$, and that $\iota_2\circ\iota_1$ maps $A$ homeomorphically to $\iota_2\circ\iota_1(A)$, with the latter considered as a subspace of $\langle C,\tau_C\rangle.$ Since $$\iota_2\circ\iota_1(A)=\iota_2\bigl(\iota_1(A)\bigr)\subseteq\iota_2(B),$$ we can through the same sort of arguments conclude that $\iota_2(B)$ is dense in $C$, and we already know that $\iota_2:B\hookrightarrow C$ is an injection, but to be able to conclude that $B$ is densely imbedded into $C$, then we also need to know that $\iota_2$ is continuous (among other things). That may have been what was meant.
A: Any inclusion of sets $X\subseteq Y$ can be seen as an injective map $\iota: X\to Y$ (simply given by $\iota(x) = x$ which makes sense since $x$ is also an element of $Y$).
The condition mentioned then is that this map is continuous, ie that we have chosen a topology on the subset such that this holds (typically, we will pick the subset topology, since this will make it hold).
A: Cameron here made ​​me realize that there are two valid answers to this question depending of which is your conceptual definition of density. 
Let $X$ be a topologycal space
First definition:

A subspace $A$ of $X$ is called dense in $X$ if the closure of $A$ respect $X$ is all of $X$.

This definition probably use your teacher. In this case we need considering a subset $A'\subset X$ equipped with a subspace topology, namely, that the inclusion map between topological spaces $i:A'\to X$ be continuous. Let $A=i(A')$ and you must show that $\overline{i(A')}=X$ (closure respect $X$).
In this case your teacher is right.
Second definition:

A subset $A$ of $X$ is called dense in $X$ if the closure of $A$ respect $X$ is all of $X$.

This definition differs from the preceding only in a conceptual fact. Given the subset $A\subset X$. You only must to show that $\overline{A}=X$ (closure respect $X$). And you are right.

In practice both definitions bring same consequences and do not change any topological facts. All topology is the same in both cases since you can always identify a subset of a topological space with the respective induced subspace topology.
A: Teacher is right. If $A\subset B$ then $i_{A\to B}\colon A\to B$ is continuous iff $A$ is subspace of $B$.
