Image of bounded operator is closed

Let $\displaystyle X, Y$ be Banach spaces, $\displaystyle U \in B(X,Y)$ be given such that $\displaystyle \exists M \in [0, \infty)$ such that $\displaystyle \forall y \in \text{Im}(U), \exists x \in X$ such that $\displaystyle U(x) = y$ and $\displaystyle \|x\|_X \leq M \|y\|_Y$. Show that $\displaystyle \text{Im}(U)$ is closed in $\displaystyle Y$.

My idea was to use the quotient Banach space of $\displaystyle \frac{X}{\text{Ker}(U)}$ and use the First Isomorphism Theorem, but this isn't very direct. Is there a more direct method for proving this? Note that $\displaystyle U$ isn't necessarily injective.