Since $X = E \oplus F$, every element in $X$ can be written uniquely as $x + y$ for some $x \in F$ and $y \in E$. Hence, we can represent an arbitrary element of $X/E$ by the coset $x + E$ for some $x \in F$. Let $T \colon X/E \rightarrow F$ be defined by $T(x+E) = x$. I claim that $T$ is an isomorphism and an isometry, where $X/E$ is given the quotient norm $\|x+E\| := \inf\{ \|x+y\| \colon y \in E \}$.
I can show that this map is an isomorphism, but I can't show that it's an isometry. In particular for any $x \in F$, since $0 \in E$ we have $\|x+E\| \leq \|x + 0\| = \|x\|$, but I cannot figure out how to show the reverse inequality.
My geometric intuition suggests we should have $\|x\| = \|x + E\|$, because we can imagine that $X = \mathbb{R}^2$ with $F$ the $x$-axis and $E$ the y-axis, and then $x + E$ is just a vertical line, and $\|x + E\|$ is the minimum distance from the line to the origin, which is just $|x|$ by the Pythagorean theorem. But, we're not even in an inner-product space, so I can't make that argument more general.
Thank you for your help.