Is given non constant continuous map is correct or not? Is the following statement true or false?
Let $f : X \rightarrow Y$ be a nonconstant continuous map of topological spaces. 
If $X$ is Hausdorff then $f(X)$ is Hausdorff.
My attempt:  I was reading this answer  but  i did not understands anything 
can anybody explain this statement: If $X$ is Hausdorff then $f(X)$ is Hausdorff...... in detail and please tell me the solution..i would be more thankful.
 A: A topological space $(X, \tau_X)$ is Hausdorff whenever every pair of distinct points have disjoint neighbourhoods. Note that this is really a condition on the topology $\tau_X$.
Take $f: X \to Y$. Define
$$ f(X) = \{ f(x) : x \in X \} \subseteq Y. $$
Whenever $f(X) = Y$ (that is, $f$ is onto), then it is clear what the topology on $f(X)$ is. But if $f(X)$ is a proper subset of $Y$, then we normally take the topology on $f(X)$ to be the relative topology
$$ \tau_f = \{ U \cap f(X) : U \in \tau_Y \}. $$
The claim (which, as noted in the comments, is not true in general) that $f(X)$ is Hausdorff whenever $X$ is Hausdorff should now be fairly self-explanatory.
To see why it is not true (even for continuous $f$), let $\tau_Y$ be the trivial topology consisting of only $Y$ and the empty set. $f$ is (trivially) continuous. Observe now that, as long as $Y$ is non-empty and not a singleton (and $f$ is not constant), $f(X)$ is not Hausdorff, even if $X$ is.
Note that $Y$ is not $T_B$ (as long as $Y$ is non-empty and not a singleton).
