Here is the key step you need to finish the proof: We are supposing $X$ is locally path-connected and semilocally simply connected, $\pi:P(X,x_0)\to \widetilde{X}$ is the quotient map identifying path-homotopy classes of paths, and $p:\widetilde{X}\to X$ is the induced surjection. Also, $ev:P(X,x_0)\to X$, $ev(\alpha)=\alpha(1)$ is endpoint evaluation.
Lemma: For any $\alpha\in P(X,x_0)$ there exists an open neighborhood $V$ of $\alpha(1)$ in $X$ and an open neighborhood $\mathcal{V}$ of $\alpha$ in $P(X,x_0)$ such that
- $ev_1(\mathcal{V})=V$,
- $\pi(\mathcal{V}\cap ev^{-1}(\alpha(1))=\{[\alpha]\}$, that is, every path $\beta\in\mathcal{V}$ that ends at $\alpha(1)$ is path-homotopic to $\alpha$.
Proof. Pick an open cover $\mathscr{U}$ of $\alpha([0,1])$ such that each $U\in\mathscr{U}$ is path connected and every loop in $U$ is null-homotopic in $X$. Using the Lebesgue number lemma, find $0=t_0<t_1<t_2<\cdots <t_n=1$ such that $\alpha([t_{i-1},t_i])\subseteq U_i$ for some $U_i\in\mathscr{U}$. In your notation, this means $\alpha\in\bigcap_{i=1}^{n}\left[[t_{i-1},t_i]\right],U_i]$. This open neighborhood of $\alpha$ is not quite good enough because the intersections $U_{i}\cap U_{i+1}$ need not be path-connected. So for each $i\in\{1,2,\dots ,n-1\}$, let $W_i$ be a path-connected neighborhood of $\alpha(t_i)$ in $U_i\cap U_{i+1}$. Now set $$\mathcal{V}=\bigcap_{i=1}^{n}\left[[t_{i-1},t_i]\right],U_i]\cap \bigcap_{i=1}^{n-1}\left[\{t_i\},W_i\right]$$ and $V=U_n$. Since $V$ is path-connected, it's clear that $ev_1(\mathcal{V})=V$. So it suffices to check that 2. holds.
Let $\beta\in \mathcal{V}$ with $\alpha(1)=\beta(1)$. It helps to draw a picture, here but basically, you want to pick paths $\gamma_i:[0,1]\to W_i$ from $\alpha(t_i)$ to $\beta(t_i)$ for each $i\in\{1,2,\dots,n-1\}$. For notational convenience, let $\gamma_0$ and $\gamma_{n}$ be the constant paths at $\alpha(0)$ and $\alpha(1)$ respectively. Now $\alpha_i$ and $\gamma_{i-1}\cdot \beta_i\cdot\gamma_{i-1}^{-}$ are paths in $U_i$ that start and end at the same point. Therefore, $\alpha_i\simeq \gamma_{i-1}\cdot \beta_i\cdot\gamma_{i}^{-}$ in $X$. By adjoining all of these path-homotopies together, you can construct a homotopy $$\alpha\simeq\prod_{i=1}^{n}\alpha_i\simeq \prod_{i=1}^{n}\gamma_{i-1}\cdot \beta_i\cdot\gamma_{i}^{-}\simeq \beta.$$Hence, if $\beta\in\mathcal{V}$ with $\beta(1)=\alpha(1)$, then $[\alpha]=[\beta]$. $\square$
Equipped with this lemma, you can finish your proof. Indeed, this is the key "technical" part of proving $\pi$ is a covering map when $\widetilde{X}$ has the natural quotient topology.
There are some interesting applications of the quotient topology on $\widetilde{X}$ when $X$ is not necessarily locally path connected or semilocally simply connected. In such cases, the quotient topology is not always equivalent to the "standard" or "whisker" topology.