According to Royden:
Mazur's Theorem: Let $K$ be a nonempty convex subset of a normed linear space $X$; $K$ is strongly closed if and only if it is weakly closed.
According to Wikipedia:
Mazur's Theorem: Let $M$ be a proper vector subspace of the topological vector space $X$ and suppose $K$ is a nonempty convex open subset of $X$ disjoint from $M$. Then there is a closed hyperplane $H\subset X$ containing $M$, but remaining disjoint from $K$.
I am not especially interested in the history or etymology of these theorems; I am really interested in whether or not they are fundamentally the same, reflections of each other. I say this because I have taken notes on the proof of Royden's version, but am currently at a loss when it comes to Wikipedia's version - if the two theorems are basically the same, I can have a go at proving Wikipedia's version myself, but if they turn out quite different it is likely that at this stage in my education I won't be able to prove Mazur's theorem (Wiki version) and I really don't want to waste my time...
Some thoughts:
For $M\subseteq H$ and $H\cap K=\varnothing$ we require a continuous $\psi\in X^\sharp$ with $H=\psi^{-1}(c)$ for some $c\in\Bbb R$: we need $\psi(m)=c$ for all $m\in M$ and $\psi(k)\neq c$ for all $k\in K$. As $M$ is convex and $K$ is open convex, one variant of the hyperplane separation theorem gives that there exists a continuous $\varphi\in X^\sharp$ such that $\varphi(k)\lt s\le\varphi(m)$ for all $k\in K,\,m\in M$ and some constant $s\in\Bbb R$, but these statements are not the same (although Wikipedia hints that there is some immediate connection). For $\varphi(m)\ge s$ for all $m$ yet we wish for $\varphi(m)=s$ for all $m$, which would require that $\varphi(m_1+m_2)=\varphi(m_1)+\varphi(m_2)=\varphi(m_1)\implies\varphi(m_2)=0$ for all $m_1,m_2\in M$, so we would like $M\subseteq\ker\varphi$ and $s=0$.
The kernel is immediately a closed hyperplane by continuity but Royden's edition of the theorem with $K$ closed is not apparently relevant.
I would say that both theorems are different faces of the same fact. At their heart, both are separation theorems.
The first theorem is typically proved the following way: If $K$ is (strongly) closed and convex and $x \in X \setminus K$, then you can find an affine closed hyperplane containing $x$ and disjoint from $K$. This implies that $K$ is weakly closed. (This also works in topological vector spaces).
In the second theorem, it should be enough to consider the case $M = \{0\}$ (by going to a quotient space). Thus you want to separate the point $0$ from the open convex set $K \not\ni 0$ by a closed hyperplane.
Now, we can relate both theorems:
If we believe in the second theorem, and $x \not\in K$ with $K$ closed and convex, then we can find an open, convex superset of $K$ not containing $x$. Thus, we can invoke the second theorem and $x$ can be separated from $K$.
Let us replace the first theorem by: "We can separate closed convex sets $C$ from points $x \not\in C$" (which is the typical way to prove theorem 1; however, I do not see immediately whether this is implied by theorem 1). This implies the second theorem: If $K$ is closed and open, $0 \not\in K$, we could consider the closed convex sets $$ C_n := \{x \in X \mid \forall \lVert{h}\rVert < 1/n : x + h \in K\} .$$ These can be separated from $0$ by some continuous linear $\psi_n$, $\lVert\psi_n\rVert = 1$. Then one can check that weak-$\star$ accumulation points of $\psi_n$ do not vanish and separate $K$ from $0$.