Why Does the DFT Assume the Transformed Signal Is Periodic?
There are already some good answers, but I still feel like adding yet another explanation, because I consider this topic extremely important for the understanding of many aspects of digital signal processing.

First of all it is important to understand that the DFT does not 'assume' periodicity of the signal to be transformed. The DFT is simply applied to a finite signal of length $N$ and the corresponding DFT coefficients are defined by

$$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N},\quad k=0,1,\ldots,N-1\tag{1}$$

From (1) it is obvious that only samples of $x[n]$ in the interval $[0,N-1]$ are considered, so no periodicity is assumed. On the other hand, the coefficients $X[k]$ can be interpreted as Fourier coefficients of the periodic continuation of the signal $x[n]$. This can be seen from the inverse transform

$$x[n]=\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}\tag{2}$$

which computes $x[n]$ correctly in the interval $[0,N-1]$, but it also computes its periodic continuation outside this interval because the right-hand side of (2) is periodic with period $N$. This property is inherent in the definition of the DFT, but it need not bother us because normally we're only interested in the interval $[0,N-1]$.

Considering the DTFT of $x[n]$

$$X(\omega)=\sum_{n=-\infty}^{\infty}x[n]e^{-jn\omega}\tag{3}$$

we can see by comparing (3) with (1), that if $x[n]$ is a finite sequence in the interval $[0,N-1]$, the DFT coefficients $X[k]$ are samples of the DTFT $X(\omega)$:

$$X[k]=X(2\pi k/N)\tag{4}$$

So one use of the DFT (but certainly not the only one) is to compute samples of the DTFT. But this only works if the signal to be analyzed is of finite length. Usually this finite length signal is constructed by windowing a longer signal. And it is this windowing which causes spectral leakage.

As a last remark, note that the DTFT of the periodic continuation $\tilde{x}[n]$ of the finite sequence $x[n]$ can be expressed in terms of the DFT coefficients of $x[n]$:

$$\tilde{x}[n]=\sum_{k=-\infty}^{\infty}x[n-kN]\tag{5}$$
$$\tilde{X}(\omega)=\frac{2\pi}{N}\sum_{k=-\infty}^{\infty}X[k]\delta(\omega-2\pi k/N)\tag{6}$$

EDIT: The fact that $\tilde{x}[n]$ and $\tilde{X}(\omega)$ given above are a DTFT transform pair can be shown as follows. First note that the DTFT of a discrete time impulse comb is a Dirac comb:

$$\sum_{k=-\infty}^{\infty}\delta[n-kN]\Longleftrightarrow\frac{2\pi}{N}\sum_{k=-\infty}^{\infty}\delta(\omega-2\pi k/N)\tag{7}$$

The sequence $\tilde{x}[n]$ can be written as the convolution of $x[n]$ with an impulse comb:

$$\tilde{x}[n]=x[n]\star \sum_{k=-\infty}^{\infty}\delta[n-kN]\tag{8}$$

Since convolution corresponds to multiplication in the DTFT domain, the DTFT $\tilde{X}(\omega)$ of $\tilde{x}[n]$ is given by the multiplication of $X(\omega)$ with a Dirac comb:

$$\begin{align}\tilde{X}(\omega)&amp;=X(\omega)\cdot\frac{2\pi}{N}\sum_{k=-\infty}^{\infty}\delta(\omega-2\pi k/N)\\&amp;=\frac{2\pi}{N}\sum_{k=-\infty}^{\infty}X(2\pi k/N)\delta(\omega-2\pi k/N)\end{align}\tag{9}$$

Combining $(9)$ with $(4)$ establishes the result $(6)$.