As a preliminary I introduce the definition of denseness I am using:
Definition (dense subsets of metric spaces). Suppose $(M,d)$ is a metric space. A subset $S \subset M$ is called dense in $M$ if for every $\epsilon > 0$ and $x \in M$ there is some $s \in S$ such that $d(x,s) < \epsilon$.
Moreover, by separable space I mean a space that admits a subset that is both dense and countable.
My aim is to prove the following claim:
Claim. Let $X,Y$ be arbitrary normed vector spaces such that $X \hookrightarrow Y$, $X$ is separable and $Y$ is non-separable. Then, $X$ isn't dense in $Y$.
My attempt. Since $X$ is separable, there is some subset of $X$ (say $A$) that is both dense in $X$ and countable. Now let us assume that $X$ is dense in $Y$ and let's try to reach a contradiction. To do this, we will try to show that $A$ is dense in $Y$ (if we indeed show that $A$ is dense in $Y$ we will be automatically showing that $Y$ is separable, which is a contradiction).
With this in mind, consider arbitrary elements $\epsilon > 0$ and $y \in Y$. Since $X$ is dense in $Y$, there exists an element $x \in X$ such that
$$ \| y - x \|_Y < \frac{\epsilon}{2}. $$
Moreover, since $A$ is dense in $X$ we also know that for this element $x$ there exists an element $a \in A$ such that
$$ \| x - a \|_X < \frac{\epsilon}{2c}, $$
where $c > 0$ is a constant that will be defined below. Therefore, we obtain
$$ \| y - a \|_Y \leqslant \| y - x \|_Y + \|x - a \|_Y. $$
Furthermore, since $X \hookrightarrow Y$, we have that
$$ \| x - a \|_Y \leqslant c \| x - a \|_X, $$
for some $c > 0$. Hence, we obtain
$$ \| y - a \|_Y \leqslant \| y - x \|_Y + c \| x - a \|_X < \frac{\epsilon}{2} + c \frac{\epsilon}{2c} = \epsilon,$$
proving that $A$ is dense in $Y$, which finishes the proof.
Thanks for any help in advance.
Your argument is completely fine, I don't see any mistake.
However, since you didn't use the non-separability of $Y$, it would be slightly cleaner to omit it and prove your result by contrapositive rather than by contradiction: indeed, you essentially proved "$X$ dense in $Y$ implies $Y$ separable", which gives by contrapositive what you desired.