Prove that $\lfloor\lfloor x/2 \rfloor / 2 \rfloor = \lfloor x/4 \rfloor$

This is straightforward using the universal property of the floor function, viz.

$\qquad\qquad\qquad\quad\, \rm k\le \lfloor r \rfloor\ \color{#c00}{\iff}\ k\le r,\ \ \ for\ \ \ k\in \mathbb Z,\ r\in \mathbb R$

$\rm\begin{eqnarray} \rm so\ for\,\ 0 < n\in \mathbb Z,\ r\in \mathbb R\!:\qquad\ &\rm k &\le&\!\rm\ \color{#0a0}{\lfloor \lfloor r \rfloor / n\rfloor} \\ \color{#c00}\iff& \rm k &\le&\ \ \rm \lfloor r \rfloor / n \\ \iff& \rm nk &\le&\ \ \rm \lfloor r \rfloor \\ \color{#c00}\iff& \rm nk &\le&\ \ \rm\ \, r \\ \iff& \rm k &\le&\ \ \rm\ \, r/n \\ \color{#c00}\iff& \rm k &\le&\ \ \rm \color{blue}{\lfloor r/n \rfloor} \\[.8em] {\rm thus}\ \ \ \rm \color{#0a0}{\lfloor \lfloor r}&\rm\color{#0a0}{ \rfloor / n\rfloor}\ &=&\rm\ \ \color{blue}{\lfloor r/n\rfloor} \end{eqnarray}$

Your problem is simply the special case $\rm\,\ r = x/2,\,\ n = 2.$

Suppose $a = \lfloor x/4 \rfloor$. Then, $a \leq x/4 < a+1$, or, $2a \leq x/2 < 2(a+1)$. As a result, $\lfloor x/2 \rfloor \in \{2a,2a+1\}\implies \lfloor x/2 \rfloor/2 \in \{a,a+\frac{1}{2}\}\implies \lfloor \lfloor x/2 \rfloor/2 \rfloor = a$.

Conversely, assume $\lfloor \lfloor x/2 \rfloor/2 \rfloor = a$. Then, $a \leq \lfloor x/2 \rfloor/2 < a+1 \implies 2a \leq \lfloor x/2 \rfloor < 2a+2$. This gives us $\lfloor x/2 \rfloor \in \{2a, 2a+1\} \implies x/2 \in [2a,2a+2) \implies x/4 \in [a,a+1) \implies \lfloor x/4 \rfloor = a$.

$\mathbf{Edit:}$ OK, I have overdone it, and one of the paragraphs will be good enough. For example, using the first one (for any given $a\in\mathbb{Z}$), \begin{align} a = \lfloor x/4 \rfloor & \iff a \leq x/4 < a+1 \\ & \iff 2a \leq x/2 < 2(a+1) \\ & \iff \lfloor x/2 \rfloor \in \{2a,2a+1\} \\ & \iff \lfloor x/2 \rfloor/2 \in \{a,a+\frac{1}{2}\} \\ & \iff \lfloor \lfloor x/2 \rfloor/2 \rfloor = a \end{align}

The suggestion that you should write $x = 4n + r$ works pretty well, where $n$ is an integer and $0 \leq r < 4$. Here $\lfloor x/4 \rfloor = n$, so you need to show that $$\lfloor \lfloor 2n + {r \over 2} \rfloor /2 \rfloor = n$$ Split into cases $0 \leq r < 2$ and $2 \leq r < 4$, or equivalently $0 \leq r/2 < 1$ and $1 \leq r/2 < 2$; it should not be that hard to verify the above equality holds in each case.

First observe that $$ \lfloor x/4 \rfloor=k\quad \text{if and only if}\quad 4k+4>x\ge 4k. $$ Next $$ \lfloor\lfloor x/2\rfloor/2\rfloor =k \quad \text{if and only if}\quad 2k+2>\lfloor x/2\rfloor\ge 2k. $$ But $$ 2k+2>\lfloor x/2\rfloor\ge 2k\quad \text{if and only if}\quad \lfloor x/2\rfloor=2k\quad\text{or}\quad 2k+1. $$ In the first case $$ \lfloor x/2\rfloor=2k\quad \text{if and only if}\quad 4k+2>x\ge 4k, $$ while in the second case $$ \lfloor x/2\rfloor=2k+1\quad \text{if and only if}\quad 4k+4>x\ge 4k+2. $$ Altogether $$ 2k+2>\lfloor x/2\rfloor\ge 2k\quad \text{if and only if}\quad 4k+4>x\ge 4k. $$

Therefore, $\lfloor x/4 \rfloor=\lfloor\lfloor x/2\rfloor/2\rfloor$.