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Starting from $\vec{\nabla}\times\vec{E} = -\partial \vec{B}/\partial t$, take a surface integral of both sides to find

\begin{align} \iint_{S(t)} \vec{\nabla}\times\vec{E}\cdot d\vec{S} &= \iint_{S(t)} -\frac{\partial \vec{B}}{\partial t}\cdot d\vec{S}\\ \therefore\quad \oint_{\partial S(t)} \vec{E}\cdot d\vec{l} &= -\frac{d}{dt} \iint_{S(t)} \vec{B}\cdot \hat{S}\, dS - \oint_{\partial S(t)}\vec{v}\times\vec{B}\cdot d\vec{l} \end{align}

where the LHS of the second line makes use of Stokes' theorem, while the RHS makes use of the Leibniz integral rule.

$\vec{B}\cdot \hat{S}=-B$ (magnetic field points into the page, while take area vector to point out) and $\iint_S dS = A(t) = l x(t)$ is the area of the (rectangular) surface enclosed by the loop, where $x(t)$ is the length of the sides that is varying with time.

So:

\begin{equation} \mathcal{E} = \oint_{\partial S(t)} (\vec{E}+\vec{v}\times\vec{B})\cdot d\vec{l} = -\frac{d(-Blx(t))}{dt} = Bl\frac{dx(t)}{dt} = Blv \end{equation}

Starting from $\vec{\nabla}\times\vec{E} = -\partial \vec{B}/\partial t$, take a surface integral of both sides to find

\begin{align} \iint_{S(t)} \vec{\nabla}\times\vec{E}\cdot d\vec{S} &= \iint_{S(t)} -\frac{\partial \vec{B}}{\partial t}\cdot d\vec{S}\\ \therefore\quad \oint_{\partial S(t)} \vec{E}\cdot d\vec{l} &= -\frac{d}{dt} \iint_{S(t)} \vec{B}\cdot \hat{S}\, dS - \oint_{\partial S(t)}\vec{v}\times\vec{B}\cdot d\vec{l} \end{align}

where the LHS of the second line makes use of Stokes' theorem, while the RHS makes use of the Leibniz integral rule.

$\vec{B}\cdot \hat{S}=-B$ (magnetic field points into the page, while take area vector to point out) and $\iint_S dS = A(t) = l x(t)$ is the area of the (rectangular) surface enclosed by the loop, where $x(t)$ is the length of the sides that is varying with time.

So:

\begin{equation} \mathcal{E} = \oint_{\partial S(t)} (\vec{E}+\vec{v}\times\vec{B})\cdot d\vec{l} = -\frac{d(-Blx(t))}{dt} = Bl\frac{dx(t)}{dt} = Blv \end{equation}

N.B. emf is defined as the work done per unit charge, \begin{equation} \mathcal{E} = \frac{dW}{dq} = \frac{d}{dq}\int \vec{F}\cdot d\vec{l}=\frac{d}{dq}\int q(\vec{E}+\vec{v}\times\vec{B})\cdot d\vec{l} =\int (\vec{E}+\vec{v}\times\vec{B})\cdot d\vec{l} \end{equation} So as Puk has pointed out in the comments, this relies on knowing the Lorentz force. See e.g. Wikipedia for a discussion.

Starting from $\vec{\nabla}\times\vec{E} = -\partial \vec{B}/\partial t$, take a surface integral of both sides to find

\begin{align} \iint_{S(t)} \vec{\nabla}\times\vec{E}\cdot d\vec{S} &= \iint_{S(t)} -\frac{\partial \vec{B}}{\partial t}\cdot d\vec{S}\\ \therefore\quad \oint_{\partial S(t)} \vec{E}\cdot d\vec{l} &= -\frac{d}{dt} \iint_{S(t)} \vec{B}\cdot \hat{S}\, dS - \oint_{\partial S(t)}\vec{v}\times\vec{B}\cdot d\vec{S} \end{align}

where the LHS of the second line makes use of Stokes' theorem, while the RHS makes use of the Leibniz integral rule.

$\vec{B}\cdot \hat{S}=-B$ (magnetic field points into the page, while take area vector to point out) and $\iint_S dS = A(t) = l x(t)$ is the area of the (rectangular) surface enclosed by the loop, where $x(t)$ is the length of the sides that is varying with time.

So:

\begin{equation} \mathcal{E} = \oint_{\partial S(t)} (\vec{E}+\vec{v}\times\vec{B})\cdot d\vec{l} = -\frac{d(-Blx(t))}{dt} = Bl\frac{dx(t)}{dt} = Blv \end{equation}

Starting from $\vec{\nabla}\times\vec{E} = -\partial \vec{B}/\partial t$, take a surface integral of both sides to find

\begin{align} \iint_{S(t)} \vec{\nabla}\times\vec{E}\cdot d\vec{S} &= \iint_{S(t)} -\frac{\partial \vec{B}}{\partial t}\cdot d\vec{S}\\ \therefore\quad \oint_{\partial S(t)} \vec{E}\cdot d\vec{l} &= -\frac{d}{dt} \iint_{S(t)} \vec{B}\cdot \hat{S}\, dS - \oint_{\partial S(t)}\vec{v}\times\vec{B}\cdot d\vec{l} \end{align}

where the LHS of the second line makes use of Stokes' theorem, while the RHS makes use of the Leibniz integral rule.

$\vec{B}\cdot \hat{S}=-B$ (magnetic field points into the page, while take area vector to point out) and $\iint_S dS = A(t) = l x(t)$ is the area of the (rectangular) surface enclosed by the loop, where $x(t)$ is the length of the sides that is varying with time.

So:

\begin{equation} \mathcal{E} = \oint_{\partial S(t)} (\vec{E}+\vec{v}\times\vec{B})\cdot d\vec{l} = -\frac{d(-Blx(t))}{dt} = Bl\frac{dx(t)}{dt} = Blv \end{equation}

Starting from $\vec{\nabla}\times\vec{E} = -\partial \vec{B}/\partial t$, take a surface integral of both sides to find

\begin{align} \iint_S \vec{\nabla}\times\vec{E}\cdot d\vec{S} &= \iint_S -\frac{\partial \vec{B}}{\partial t}\cdot d\vec{S}\\ \therefore\quad \oint_{\partial S} \vec{E}\cdot d\vec{l} &= -\frac{d}{dt} \iint_S \vec{B}\cdot \hat{S}\, dS \end{align}

where the second line makes use of Stokes' theorem.

$\vec{B}\cdot \hat{S}=-B$ (magnetic field points into the page, while take area vector to point out) and $\iint_S dS = A(t) = l x(t)$ is the area of the (rectangular) surface enclosed by the loop, where $x(t)$ is the length of the sides that is varying with time.

So:

\begin{equation} \mathcal{E} = \oint_{\partial S} \vec{E}\cdot d\vec{l} = -\frac{d(-Blx(t))}{dt} = Bl\frac{dx(t)}{dt} = Blv \end{equation}

Starting from $\vec{\nabla}\times\vec{E} = -\partial \vec{B}/\partial t$, take a surface integral of both sides to find

\begin{align} \iint_{S(t)} \vec{\nabla}\times\vec{E}\cdot d\vec{S} &= \iint_{S(t)} -\frac{\partial \vec{B}}{\partial t}\cdot d\vec{S}\\ \therefore\quad \oint_{\partial S(t)} \vec{E}\cdot d\vec{l} &= -\frac{d}{dt} \iint_{S(t)} \vec{B}\cdot \hat{S}\, dS - \oint_{\partial S(t)}\vec{v}\times\vec{B}\cdot d\vec{S} \end{align}

where the LHS of the second line makes use of Stokes' theorem, while the RHS makes use of the Leibniz integral rule.

$\vec{B}\cdot \hat{S}=-B$ (magnetic field points into the page, while take area vector to point out) and $\iint_S dS = A(t) = l x(t)$ is the area of the (rectangular) surface enclosed by the loop, where $x(t)$ is the length of the sides that is varying with time.

So:

\begin{equation} \mathcal{E} = \oint_{\partial S(t)} (\vec{E}+\vec{v}\times\vec{B})\cdot d\vec{l} = -\frac{d(-Blx(t))}{dt} = Bl\frac{dx(t)}{dt} = Blv \end{equation}