The entire content is rather drafty, but I am especially baffled with the last comment "and thus produces an image manifold that is a diffeomorphic copy of $X$ adjacent to the original." This sentence does not make sense to me, like why we suddenly discuss original, why it is this case, and why this matter?
$\quad$We must deal with the necessity of deforming $X$ in a mathematically precise manner. Attempting to define deformations of arbitrary point sets in $Y$ is hopeless, so we shift our point of view somewhat. Considering $X$ as an abstract manifold and its inclusion mapping $i:X\hookrightarrow Y$ simply as an embedding, we know how to deform $i$, namely by homotopy. Since embeddings form a stable class of mappings, any small homotopy of $i$ gives us another embedding $X\to Y$ and thus produces an image manifold that is a diffeomorphic copy of $X$ adjacent to the original.
If $f:X\to Y$ is a smooth embedding, then the image $f(X)$ is diffeomorphic to $X$.
Let $F:X\times (-\epsilon,\epsilon)\to Y$ be a smooth map, and write $f_t(x)$ for $F(x,t)$. Thus, we have a family of smooth maps $f_t:X\to Y$. Figuratively speaking, when $t,s\in (-\epsilon,\epsilon)$ are "close", the images $f_t(X)$ and $f_s(X)$ can be said to be "adjacent". Consider this gif of a homotopy (from wikipedia):
For times $t$ and $s$ that are close together, the image curves at time $t$ and time $s$ are "adjacent".
"Embeddings form a stable class of mappings" means that if $f_t$ is an embedding for some $t$, then for some neighborhood $S\subseteq(-\epsilon,\epsilon)$ of $t$, we will have that $f_s$ is an embedding for all $s\in S$.
Thus, if $X\subseteq Y$ is an embedded submanifold of $Y$ (so that $i:X\hookrightarrow Y$ is a smooth embedding), then a homotopy of $i$ (that is, a map $F:X\times(-\epsilon,\epsilon)\to Y$ with $f_0=i$) will, for $t$ sufficiently close to $0$, produce embeddings $f_t:X\to Y$, whose image manifolds $f_t(X)$ are "adjacent" to $X$ and diffeomorphic to $X$.