Let us consider only Banach spaces.
A classical theorem of Riesz says that for every proper closed subspace $Y$ of a Banach space $X$ there is a sequence of vectors $x_n\in X$ with $\Vert x_n\Vert=1$ for all $n$ and such that $\Vert x_n+Y\Vert\to 1$, where this latter norm is the usual norm in the quotient space $X/\!Y$: $$\Vert x+Y\Vert=\inf_{y\in Y}\Vert x+y\Vert.$$
I will call the validity of the natural strengthening of this theorem the strong Riesz property (SRP):
A Banach space has the SRP iff for every proper closed subspace $Y$ of $X$, there is $x\in X$ with $\Vert x\Vert=\Vert x+Y\Vert=1$.
Two applications of Hahn-Banach (on $X$ and on the dual $X^*$) imply that every reflexive space has the SRP.
A closely related property is given by James' theorem, which I will also expres as a property:
A Banach space $X$ has James' property (JP) iff for every continuous linear functional $f\in X^*$ attains its norm at some unit vector; i.e., there exists $x\in X$ with $\Vert x\Vert=1$ such that $|f(x)|=1$.
James' theorem states that JP and reflexivity are equivalent.
I would like to know whether reflexivity and the SRP are equivalent. It seems that passing through JP would be a natural way to prove it.
The implications (reflexivity)$\Rightarrow$(JP)$\Rightarrow$(SRP) are easy. For example, the last one: Let $Y$ be a proper closed subspace of a Banach space with JP. By Hahn-Banach, take $f\in X^*$ with $f(Y)=0$ and $\Vert f\Vert=1$. By JP, take $x\in X$ with $\Vert x\Vert=|f(x)|=1$. Then $$\Vert x+Y\Vert\leq\Vert x\Vert=1=|f(x)|=\inf_y|f(x+y)|\leq\inf_y\Vert x+y\Vert=\Vert x+Y\Vert,$$ which yields SRP.
The "obvious way" to prove the converse would be: Assume $f\in X^*$ has norm $1$. By SRP, take $x\in X$ with $\Vert x\Vert=\Vert x+\ker(f)\Vert=1$. Then proceed to prove that $|f(x)|=1$.
However, this very last step does not seem to be valid, or at least not in any easy manner. Everything that I could obtain is that $|f(x)|\leq 1$.
tl;dr Does SRP as above imply reflexivity?
I suppose the OP forgot to require that $\|f\|=1$, above. Assuming this, one has for every $y$ in $X$ that $$ k:=f(x)y-f(y)x\in \ker(f). $$ Therefore $$ |f(y)| = |f(y)| \|x+\ker(f)\| = \|f(y)x+\ker(f) \| = $$$$ = \|f(x)y+\ker(f)\| \leq |f(x)|\|y\|. $$ So, $$ 1 = \|f\| = \sup_{\|y\|\leq 1}|f(y)| \leq |f(x)| \leq \|f\|\|x\| = 1. $$