Recall that for a given continuous map $f:X\to Y$, the mapping cylinder of $f$ is defined as $M_f=((X\times I)\sqcup Y)/\sim$ where $(x,t)\sim y$ if $y=f(x)$ and $t=1$.
Let us define $i:X\to M_f$ by $i(x)=[(x,0)]$. We have to show that $i$ is an embedding.
It's easy to see that $i$ is one-one as for $x\ne x'$, $(x,0)\not\sim (x',0)$.
Also observe that $i$ is continuous as $i=\gamma\circ g$ where $g:X\to (X\times I)\sqcup Y$ defined as $g(x)=(x,0)$ and $\gamma:(X\times I)\sqcup Y\to M_f$ is the usual identification map.
Here $\gamma$ is by definition continuous and $g$ is an embedding. Hence, $i$ is continuous.
Only thing that remains to show for any open set $U\subset X$, $i(U)$ is open in $\text{Im}(i)\subset M_f$. I couldn't write that part properly. Can anyone help me to complete the proof?
Here is a hint:
Suppose, $h:X\times I\to(X\times I)\sqcup Y$ be the natural inclusion. Now try to prove that, $\gamma\circ h(U\times [0,1))$ is open in $M_f$ and then, show that, $i(U)=\gamma\circ h(U\times [0,1))\cap \text{Im}(i)$