Showing that $0 \in \overline{A(\partial B_1^X)}$ if $X,Y$ Banach with $\dim X = + \infty$ and $A \in \mathcal{L}_c(X,Y)$. Exercise :

Let $X,Y$ be Banach spaces with $\dim X = + \infty$ and $A \in \mathcal{L}_c(X,Y)$. Show that $0 \in \overline{A(\partial B_1^X)}$.

Attempt :
So, since $A$ is a Linear Compact operator $A : X \to Y$, this means that it "transfers" bounded subsets of $X$ to relatively compact sets in $Y$ (meaning their closure is compact). The unit ball $B_1^X$ in $X$ is such a bounded set. That means that $\overline{A(B_1^X)}$ is a compact set. 
Now, since $\partial B_1^X = \{ x \in X : \|x\|_X = 1\}$ does that mean that it is also bounded ? 
Then, if that's correct, $\overline{A(\partial B_1^X)}$ should be compact.
Since $A \in \mathcal{L}_c(X,Y)$ then the image (range) of $A$ is separable, but I don't know (or see) how I could get that into the mix here.
How would one actually show that $0 \in \overline{A(\partial B_1^X)}$ ?
 A: We want to show that there is a sequence of unit vectors $(x_n)_{n\ge 1}$ such that $\lim_{n\to\infty}Ax_n=0$. Note that this is equivalent to $0\in \overline{A(\partial B_1^X)}$. Assume to the contrary that there exists a $c>0$ such that for every unit vector $x$, $\|Ax\|\ge c$, or equivalently, $\|Ax\|\ge c\|x\|(*)$ for all $x\in X$. Then it follows that $A$ is injective and has a closed range. This is because if $Ax_n\to y$, then $\|A(x_n-x_m)\|\to 0$ as $n,m\to\infty$, which in turn implies that $(x_n)_{n\ge 1}$ is Cauchy by $(*)$. Thus it follows that $A:X\to A(X)$ is a bijection between two Banach spaces, so there exists a continuous inverse $A^{-1}$ by inverse mapping theorem. Now, observe that $I_X=A^{-1}A:X\to X$ should be a compact operator as it is a product of a compact operator and some bounded linear operator. However, since $X$ is infinite-dimensional, $I_X$ cannot be compact. This leads to a contradiction. So there is $(x_n)_{n\ge 1}\subset B_1^X$ such that $Ax_n \to 0$ as $n\to\infty$.
