Let $\mathcal{f}: \mathbb{R} \to \mathbb{R}$ be a continuous function and consider $\mathcal{F}=\lbrace (x,y)\in \mathbb{R}^2 \mid y>f(x) \rbrace $.Prove that $\mathcal{F}$ is path-connected.
Proof
Let $P=(p_1,q_1),Q=(p_2,q_2)$ be points of $\mathcal{F}$ we should be find a path between $P,$ and $Q$.For $\varepsilon > 0$ Consider $K=(x,f(x)+\varepsilon)\in \mathcal{F}$.
Let $\alpha_{1}:[0,1] \longrightarrow \mathcal{F}$, with $\alpha_{1}(t)=(p_1+t(x-p_1),q_1+t(f(x)+\varepsilon -q_1))$ which connect $P$ with $(x,f(x)+\varepsilon)$, notice that The path is always well defined since $\mathcal{f}$ is continuous function and $d\left((x,f(x))(x,f(x)+\varepsilon))\right)<2 \varepsilon $ and in fact $\alpha_1$ is continuous since is a linear function of $t$.
Of the same way define $\alpha_{2}:[0,1] \longrightarrow \mathcal{F}$, given by $\alpha_{2}(t)=(x+t(p_2-x),f(x)+\varepsilon+t(q_2-f(x)+\varepsilon))$ wich connect $(x,f(x)+\varepsilon)$ and $Q$.
Now consider $\alpha[0,1]\longrightarrow \mathcal{F}$ given by $\alpha(t)=\begin{cases} \alpha_{1}(2t) \, \, \text{ If $t\in[0,\frac{1}{2}$]}& \\ \alpha_2(2t-1) \, \, \text{If $t\in [\frac{1}{2},1]$} \end{cases}$ Which is continuous by pasting lemma and connect $P$ to $Q$
Hence $\mathcal{F}$ is Path-connected
Let $M$ be the maximum value of $f$ on $[p_1,p_2]$. Then the combination of the vertical line segments from $P$ to $(p_1, M+1)$, from $(p_1, M+1)$ to $(p_2,M+1)$ and then down from $(p_2, M+1)$ to $Q$ is the required path.
No weird $\varepsilon$'s or anything. Just go up, it's a safe way to stay inside $\mathcal{F}$.