Polynomial Time reducible explanation
What is polynomial-time reduction?
Let us recall the definition of polynomial-time reduction (function) as presented in section 34.3, NP-completeness and reducibility
CLRS, third edition.

Let $\Sigma=\{0,1\}$. A language $L_1$ over $\Sigma$ is polynomial-time reducible to a language $L_2$ over $\Sigma$, written $L_1 \le_p L_2$, if there exists a polynomial-time computable function $$f: \{0, 1\}^*\to\{0,1\}^*$$ such that for all $x\in \{0,1\}^*$,
$$x\in L_1\Longleftrightarrow f(x)\in L_2.$$ We call the function $f$ the reduction (function), and a polynomial-time algorithm $F$ that computes $f$ is a reduction (algorithm).


Examples

[FALSE]$\ \ $ If set $Y$ can be solved in $O(2^n)$ and $Y \leq_p X$ then $X \not\in \bf{P}$.

What is $O(2^n)$? It is not $\Omega(2^n)$! For example, a constant function is in $O(2^n)$. If set $Y$ can be solved in constant time and $Y \leq_p X$ then $X$ can be in $\bf{P}$. For example, we can let $X=Y$.

[FALSE]$\ \ $ If $Y\le_p X$ and $X$ is solvable in constant time, then $Y$ is solvable in constant time as well.

You cannot ignore the time spent in the reduction step. $Y\le_p X$ tells us that it takes no more than polynomial time to map (a.k.a. reduce) an instance (a.k.a a word) of $Y$ to an instance of $X$. That polynomial time, as you suspected, is significant.
In fact, the complexity class $\mathbf P$ can be defined as all $Y$s such that $Y\le_p X$ for some $X$ that is solvable by constant time.

[ALMOST TRUE]$\ \ $ Say you have $SORT$ which checks if the list of ints is sorted, then for all $X$ we have $SORT \leq_p X.$

There is a polynomial-time computable function $f$ that maps a word in $\{0,1\}^*$ that encodes a sorted list of ints to 1. Basically, we can use an algorithm that checks each int in the list is not greater than than the next int in the list. This algorithm is, in fact, linear-time.
Suppose word $x_0\not\in X$ and $x_1\in X$. Let $g$ be a function that maps 0 to $x_0$ and 1 to $x_1$. $g$ can be computed by a constant-time algorithm, since both $x_0$ and $x_1$ are fixed sequences of 0s and 1s. $g\odot f$ is a polynomial reduction from $SORT$ to $X$.
As long as $X$ is not the empty set nor the set of all words over $\Sigma$, $SORT \leq_p X.$
If $X$ is empty set, there is no reduction of any kind. If $X$ is the set of all words, then any map to $X$ will map any instance, including a &quot;no&quot; instance to a &quot;yes&quot; instance of $X$. Because of these two exceptions, we can say the proposition is false.

[TRUE]$\ \ $ Assume $Y \leq_p X$, if $Y$ doesn't have a polynomial time algorithm to solve it, then there isn't one for $X$ either

$Y \leq_p X$ implies
$\quad X$ is solvable in polynomial time $\Longrightarrow$ $Y$ is solvable in polynomial time.
Note that contrapositive of that implication is exactly
$\quad Y$ is not solvable in polynomial time $\Longrightarrow$ $X$ is not solvable in polynomial time.
Exercises
Check whether the following propositions are true.
Proposition 1.  If set $Y$ can be solved in $\Omega(2^n)$ and $Y \leq_p X$ then $X \not\in \bf{P}$.
Proposition 2.  If set $X$ can be solved in $O(2^n)$ and $Y \leq_p X$ then $Y \not\in \bf{P}$.
Proposition 3.  If set $X$ is solvable by polynomial time, then $\mathbf P$ can be defined as all $Y$s such that $Y\le_p X$.